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Semi-circles and right angle dilemmas . . .

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Daily Mail reports on a class assignment for seven year olds that happened to be set for the daughter of a Mathematics Lecturer at Oxford.

Maths lecturer is left baffled by his seven-year-old daughter’s geometry homework and turns to Twitter for help – so can YOU work out if it’s true or false?

Dr Kit Yates shares his seven-year-old daughter’s maths homework to Twitter

The question asked students whether a semi-circle had ‘two right angles’ or not

The maths lecturer, from Oxford, admitted that he was stumped by the problem 

People were left baffled by the question and came up with conflicting answers 

By Kate Dennett For Mailonline

Published: 17:40 GMT, 25 February 2021 | Updated: 17:40 GMT, 25 February 2021

Here is the question and the “expected” seven year old School Math answer:

Why would a PhD level Maths Lecturer disagree? Indeed, why would he suggest — and this is where it gets interesting for us here at UD — “there was a strong case for claiming that the answer was ‘true’ as well as ‘false’“?

Sounds familiar?

Obviously, a Math Lecturer — I switch to the American style abbreviation — will be familiar with the world of the infinitesimal. So, he would be aware that at the point of intersection of diameter and circumference, there is a required right angle to the tangent and that at that point curve and tangent coincide . . . the tangent and the curve share the same slope at that point, but not at any identifiable real value on the circumference beyond — thus, distinct from — that point.

Arguably, for an infinitesimal spread about the point where the tangent touches the curve and the diameter cuts it, we have a straight line segment c_0 +/- dc in the circumference but at right angles to the diameter. That is, the hyperreals are peeking in and giving us a language to constructively talk about structures and quantities.

Let’s cross check, through Wikipedia:

In geometry, a polygon (/ˈpɒlɪɡɒn/) is a plane figure that is described by a finite number of straight line segments connected to form a closed polygonal chain or polygonal circuit. The solid plane region, the bounding circuit, or the two together, may be called a polygon.

The segments of a polygonal circuit are called its edges or sides, and the points where two edges meet are the polygon’s vertices (singular: vertex) or corners. The interior of a solid polygon is sometimes called its body. An n-gon is a polygon with n sides; for example, a triangle is a 3-gon.

Notice, the requirement, “finite number of straight line segments”? Other common sources will omit finiteness, and it is commonly recognised that a circle is a beyond finite case of a regular n-gon. So, we have reasonable grounds for referring to infinitesimal line segments and removing the finitude. For example, that is one way to approach the question, what is the area of a circle.

I should note that an infinitesimal value can be seen as a number say h near to 0 in the number line, closer to 0 than any 1/n, where n is a finite counting number. (This opens up also the concept of transfinite hyperreals. That is h = 1/H, a number such that H > n for any finite counting number.)

Shifting to Coordinate Geometry, let’s put our circle at the origin, O, and give it radius r:

With these ideas in hand, we can see that by using infinitesimals, a circle is a limiting case of a regular polygon, but the length of sides is too short to be captured by any distinct point on the circumference c_1 at (x1,y1) distinctly removed from C_0, at (x0,y0) such that x1 – x0 is a real value, and the same for y1 – y0. (That is, we here mark a distinction between the infinitesimally altered x0 + dx or the similarly altered y0 + dy and any neighbouring x1 and y1, constrained to be strictly real.)

This allows us to seemingly have our cake and eat it.

That is, no distinctly removed point c1 with strictly real coordinates different from c0 will lie on the tangent to the circle at c0 at the point where the diameter cuts the circle. But at the same time we can speak reasonably of points along a short line segment at c0, that is straight but in the circumference, as the points in question c0 + dc, are infinitesimally removed from c0. Where, dc^2 = dx^2 + dy^2.

That is, further, we are looking at an infinitesimal right angle triangle with the hypoteneuse dc being on the tangent line for the circle at c0.

If we can swallow this camel, then we have an answer, yes there is a right angle at the intersections of diameter with circumference, and yes the circle is bounded by a closed curve where any distinct c_1 at (x1,y1) that is such that x1 and y1 are strictly real, will fall away from the tangent at c_0.

Is such reasonable?

Well, let us see how we arrived at the differential coefficient dy/dx, as an expression of the slope of y = f(x) at some point on a curved line, through the slope of the tangent-limit to the slopes of the secants:

Here, we see how the slopes of secants, in the limit, become the slope of the tangent at M, as h tends to 0, i.e. becomes an infinitesimal.

Now, let us ponder the curvature of a continuous, differentiable line:

Here, curvature K = 1/R, R the radius of curvature. Observe that the tangent line is perpendicular to the curve and its osculating circle, which near P is coincident with the curve. All of this begs for interpretation on infinitesimals. BTW, for a straight line, the radius of curvature is infinite, yielding a curvature of 0.

With these in mind, we can see that no, this is not a superposition of two contradictory states but a case that is surprisingly relevant, regarding the value of the infinitesimal perspective.

Now, where does this all lead? Interesting places . . . END

VL, I had in mind the point of inflexion -- momentary straightness -- case, which is necessarily present when sense of curvature changes. What I learned way back specifically does not include that a tangent does not cross a curve; which a source implicitly used above seemed to say. KF kairosfocus
Yes, KF, I discussed the point of inflection case. Also, yes the fact that the tangent line has a slope given by the derivative is the best mathematical way to describe a tangent line. This covers the endpoint case and the point of inflection case. You write, "An issue arises on a contrast with what I learned, which leaves off the does not cross: tangent at an inflexion point." I'm not sure what that means. Are you saying there is some other case where a tangent line crosses the curve (a "standard" continuous differentiable curve) other than at a point of inflexion? If so, can you describe the case? Viola Lee
VL, that "tangent touches a curve at a single point (and does not cross)" is a case of a studied seemingly simple but deeply loaded remark that I think embeds a mistake on a "pathological" case. Any other real coordinates point on the curve is already falling away from the tangent. But, does such have to be on the same side? An issue arises on a contrast with what I learned, which leaves off the does not cross: tangent at an inflexion point. I suspect, touches at one point and shares the same slope there, may cover this case. Of course, on apolitical matters Wiki can be helpful:
In geometry, the tangent line (or simply tangent) to a plane curve at a given point is the straight line that "just touches" the curve at that point. Leibniz defined it as the line through a pair of infinitely close points on the curve.[1] More precisely, a straight line is said to be a tangent of a curve y = f(x) at a point x = c if the line passes through the point (c, f(c)) on the curve and has slope f'(c), where f' is the derivative of f. A similar definition applies to space curves and curves in n-dimensional Euclidean space. As it passes through the point where the tangent line and the curve meet, called the point of tangency, the tangent line is "going in the same direction" as the curve, and is thus the best straight-line approximation to the curve at that point. Similarly, the tangent plane to a surface at a given point is the plane that "just touches" the surface at that point. The concept of a tangent is one of the most fundamental notions in differential geometry and has been extensively generalized; see Tangent space. The word "tangent" comes from the Latin tangere, "to touch".
The idea of infinitely close points of course is a peeking in of infinitesimals. (Onward, transfinite hyperreals: 1/h --> H, > any finite n in N.) KF kairosfocus
KF writes, " I think teachers need to be more aware of the big issues next door to what we talk about question. It is hard to be simple, clear and building of core capabilities without occasionally going in over our heads. " This is important. That is why the remark in the picture that "There is a difference between the correct answer and the answer at this learning stage” is poor pedagogy. Viola Lee
VL & SA2, yes, that's why I thought of the tiny ant metaphor. At the corner, it pivots through an angle, in the limit a right angle. We can argue, two branched function on a range specified by the corners shared in common. There are workarounds for the one result rule. Also, the value of limits and tamed infinitesimals becomes evident. I think teachers need to be more aware of the big issues next door to what we talk about question. It is hard to be simple, clear and building of core capabilities without occasionally going in over our heads. Things may not be as simple as we have been led to believe, e.g. I recall counting by tens and getting out of depth with my second dad then a next door neighbour: . . . nine-ty . . . nine-ty nine, ten-ty. I followed the rule not knowing the scale transition, this was the first time I counted to 100, and it was of an evening after school. I was corrected, one hundred. I immediately saw, counting by hundreds as extension of counting by tens. I did not go on to ten hundred that day, but lurking there is an infinity as well as the convention-riddled way we group numbers. I think in this case we are looking at, the ant metaphor might be helpful. Somebody tried to give a counter example to an angle and got in over the head where the great whites swim. KF kairosfocus
Q, yes, all of these are feasible. The similar triangles was used to deduce things about distance to moon and sun way back. Shadow of earth on moon in an eclipse shows 3:1 ratio of diameters of shadow and object, mix in Eratosthenes' work and we have our first sol system scale yardsticks. A pity work on the Sun was marred by issues on refraction etc. My bet is, lowering on a string and similar triangles with shadows are the best methods. KF kairosfocus
KairosFocus @70, Yes, exactly. You got two of the student's options. Other solutions he suggested included - Lowering the barometer on a string and then measuring the length of the barometer plus the string. - Swinging the string to determine the period of the pendulum (the difference in G being negligible). - Measuring the lengths of the shadow of the barometer and the building and applying the ratio to the height of the barometer. - Setting up a line of sight between the top of the barometer and the top of the building and measuring the ratios of the resulting similar triangles. And a bunch more like these! You're right that barometric pressure difference is probably the poorest method of trying to measure the height of the building, a fact that eluded the teacher. Can anyone here find any other methods that the student might have listed? -Q Querius
I understand your questions, Steve, and will try to explain. I wish I could draw pictures, though. Here are several relevant points 1. We are not thinking of the semicircle and the diameter as a single curve, but rather as two separate curves. The diameter has angle of 0° in respect to the horizontal. The angle of a tangent line on the semicircle is different everyplace. At the bottom of the semicircle the tangent line is also horizontal. Image a point P starting at the bottom of the semicircle and moving to the right along the circle as it approaches the intersection with the diameter: it gets steeper all the time. At the exact moment P reaches the diameter the tangent line would be perfectly vertical, and thus perpendicular to the diameter. If we were thinking of the shape as a single function (which we are not), then the "point" where the diameter and semicircle meet would not have a single tangent line, because the value you get approaching that point along the diameter would be different than the value you get approaching along the semicircle. 2. Also, you offer this definition of a tangent line "“a straight line or plane that touches a curve or curved surface at a point, but if extended does not cross it at that point.” I see you got this from Encyclopedia.com. There are two ways this is incomplete. One is at the endpoint of a curve, which we are discussing here in respect to the semicircle. This is more obvious in respect to a smaller part of a circle. Imagine a point P at the 45° point on a circle, starting with 0° at the far right. It has a tangent at that point perpendicular to the radius at a 45° angle also. Now throw away all the circle except the part from 0° to 45°, so we just have an eighth of a circle. The tangent at 45° is still exactly what it was even though the circle doesn't continue on past that point. The second situation is what is called a point of inflection. Image the letter S. The letter curves one way at the top, and then back the other way at the bottom. Right where the curvature changes (the point of inflection) there is a tangent line that is"outside" and below the top part of the S but "outside" and above the bottom part. This is a tangent line that does cross the curve at that point. To really get a tangent line you have to use the process KF showed in the OP about turning a secant line into a tangent line using the process of a limit. Viola Lee
Steve Alten2: The tangent is undefined at corners or places where more than one straight line can go through only that point on the 'curve'. JVL
Viola Lee ” But the definition of the angle between two curves (and the line is a curve, mathematically speaking) is that it is the angle between the tangent lines.” I apologize for not quite getting this. I understand that for any point along an arc there can only be one tangent. But how would this apply to the shape displayed at the point where the arc changes to a straight line (the vertices)? Especially given the definition of a tangent. “ a straight line or plane that touches a curve or curved surface at a point, but if extended does not cross it at that point.” At the vertice, a tangent could be 90 degrees to the straight line, but it could also be greater than 90. What I am saying is that you can pivot the tangent around this point an additional 90 degrees without crossing any plane. Steve Alten2
Q, I suspect, unless you have a skyscraper, the barometer will not be an effective index of difference in height. Most obvious, time how long it takes to hit ground or use it as weight on a string. KF kairosfocus
VL, yes. I would use a glass standing in for a bucket and a lab tap, to give rates and accumulations of change as flows and stocks. Then, slopes and areas under curves, then fundamental theorem on mutually inverse operations. I would do a bell shaped pulse and show the sigmoid. That points onward to a lot of other things. KF PS: How that definition came to be settled is the root of the matter. Hint, Calculus in key part arose out of need to analyse curved motion of planets etc. kairosfocus
Educators are sometimes guilty of discouraging students from thinking outside the box. There was a famous test question in the 1950s (IIRC) that a creative student answered in an out-of-the-box way, refusing to regurgitate the obvious. The question was 1. How could you measure the height of a building using a barometer? The student suggested that there is a large number of options available and provided a list. Can you see the student's point? Oh, and the teacher didn't know whether to give the student credit and so on advice, gave the student the *same* question--a second chance to conform to "science." The student then provided more options not provided previously, none of which were the desired, obvious answer. -Q Querius
JVL, ah, i rounded from a rounding teach me to underestimate the old girl, Penny, reporting a 4-place value. KF kairosfocus
Steve, you write, "The only angles possible would be those between the straight line and tangents of the arc at the point they meets the straight line. But that is not the question that was asked." But the definition of the angle between two curves (and the line is a curve, mathematically speaking) is that it is the angle between the tangent lines. Of course the kids wouldn't know that, but it is true that the semicircle and the diameter meet at a right angle. Viola Lee
WJM, here is a thought experiment about this issue that I used in my very first introduction to the derivative in my "Intro to Calc" class. Imagine a falling ball. We know that gravity constantly accelerates a falling object, so at every moment the velocity of the ball is greater than it was the moment before. Suppose we want to measure the velocity of the ball at time t(0): call this desired number v(0). Also suppose we have extremely accurate measuring devices. We could measure the height at t(0) and the height at some other time a short while later, t(1). Then we could compute the average velocity between t(0) and t(1) by dividing the change in the height by the change in the time. Call this number v(1). Since the ball is falling faster and faster we know that v(1) is somewhat faster than v(0), so we have a least an estimate for v(0) that we know is a little too big. How could we get a better estimate? Obviously, we could measure the height at a different time t(2) that is closer to t(0). (That is, the change in time t(2) - t(0) is smaller.) This would give us a new, and better estimate, of the velocity v(0) because the moments in time were closer together. How could we do better? Pick moments in time closer and closer to t(0). However, we always have to have second point: we can't let the second point also be t(0), because then we have no change in height and no change in distance, and that doesn't produce a velocity. But we can let our second point get closer and closer to t(0), indefinitely. and what we find is that our computed average velocity gets closer and closer to a particular velocity. At some point we realize that the computed velocity approaches a limit, so we say that the limit of the computed velocity, as the time interval gets indefinitely small, is in fact the instantaneous velocity v(0) we were looking for. So even though we can't compute the instantaneous velocity directly because at a moment in time there is no change in height or change in time, we can still determine the instantaneous velocity by taking the limit formed by considering smaller and smaller intervals of time. In calculus language, the instantaneous rate of change is the limit, as change in time approaches 0, of the quotient of the change in height divided by change in time. (This is the formula in KPF's OP.) This works, and in this and countless other places both leads to a consistent mathematical system and agrees with the real world when properly applied. Obviously the ball is falling. It is not frozen in time just because we can't compute zero divided by zero. This method is the heart of calculus, without which our modern world of science and technology would not exist. Viola Lee
Let’s get back to the question that was asked of the student. I agree with WJM that there are no angles in the shape displayed. The only angles possible would be those between the straight line and tangents of the arc at the point they meets the straight line. But that is not the question that was asked. Steve Alten2
Viola Lee “ Steve, I assume you know you are talking about an entirely different problem?” Sorry, just my lame attempt at humour, playing on Kairosfocus’ frequently used cliches that you can’t square a circle. Steve Alten2
Kairosfocus: About 1.773 Not the right rounding for sqrt(pi) which is approximately 1.772453850905516 . . . JVL
SA2, yes you can use Calculus to determine the area of a circle. A = pi * r^2. Which then gives the side of a square of equivalent area, root-pi times r. About 1.773 times r. KF kairosfocus
Steve, I assume you know you are talking about an entirely different problem? Viola Lee
Some responses: First, let's be clear what we are talking about: the semicircle is one curve and the diameter is another (assuming that is what they are.) We are not talking about a semicircle in isolation nor are we talking about the picture as one single function. So, it is true that if we consider just a semicircle, there are no right angles, or angles of any sort, on the semicircle. And if we consider the shape as a single function, with the place where the diameter and the semicircle meet as P, the function is undifferentiable at P because the right and left hand derivatives are different. But if we ask clearly what angle does the semicircle make with the diameter, the answer is 90°. The reason is, and this is official calculus, the angle between two curves is the angle between their tangent lines at the point of intersection. The central issue that calculus solves is how to find the instantaneous rate of change (or as KF says, instantaneous velocity) at a moment when the rate of change is continuously changing. This is done via the concept of limit. WJM expresses the dilemma of pre-calculus understanding in 52 and JVL responds correctly at 56, "Calculus is the mathematics of change. One of the greatest intellectual achievements of the human race." We would have to have better software, and lots of time, to give WJM a beginning lesson, although KF summarized the approach in the OP in the section "From Slopes of a Secant to Slopes of a Curve." Viola Lee
See, you can square a circle. :) Steve Alten2
WJM, VL was right to point out that an object constrained to move in a circle, at each point of the arc is moving along the tangent. This connects to the issue, instantaneous velocity: at a given point, but in motion and under acceleration. In the case, release and our bung flies off along the tangent, then of course is influenced by the ballistics forces. This is how the old fashioned sling works. KF kairosfocus
William J Murray: To put it simply: a circle has no straight line segments, so when it intersects any other line, curved or straight, there is no angle produced unless one is talking about something else. This is the point of calculus: what is a function doing at a single point (in time?) even if it's changing. Calculus is the mathematics of change. One of the greatest intellectual achievements of the human race. JVL
JVL, no, there are two different slopes, approaching the corner from two different paths and pivoting at the corner. The pivoting is real, the angle of pivot is real and the angle at the corner has a legitimate meaning. All of this points to the infinitesimals and limits, thence R*. KF kairosfocus
To put it simply: a circle has no straight line segments, so when it intersects any other line, curved or straight, there is no angle produced unless one is talking about something else. William J Murray
William J Murray: the second being “right next to it” on the circle but off the dissecting line. There is no point right next to another one. Time to move on folks. JVL
Maybe calculus covers this, but I never took calculus. To me, it seem obvious that the "angle" cannot be anything other than that created by an abstract straight line using two plot points on the circle, one being where it intersects with the dissecting line and the second being "right next to it" on the circle but off the dissecting line. The one "off" the dissecting line cannot be perpendicular to the line (nature of a circle), so there cannot be a "right angle" there. But further, you're not even talking about the line of the circle any more, you're talking about a straight line plotted from two points on the circle. Not only is there not a right angle there, there isn't any angle whatsoever. There is no "angle" produced by the circle dissected by the straight line. This goes back to logic and identities. You can't measure that angle because no measurable angle is produced; you have to be measuring the angle of something else. As it turns out, what you're measuring is the angle generated by two intersecting straight lines. William J Murray
Kairosfocus: the shift in slope at the corners gives an angle as the ant metaphor will showk The slope approaching the corner from the two 'sides' gives two different answers. Therefore, the slope at the 'corner' is undefined. Basic calculus stuff. Time to move on. JVL
JVL, kindly see the just above. KF kairosfocus
VL, yes, imperfect curriculum, and yes the Calculus, is the key answer. Yes, telling a misleading answer to the young is problematic. The next bit is, there is a reasonable sense that there is a slope of a curved line, tied to the slope of its tangent. at any point a rubber bung on a string is moving along the tangent and sees a centripetal acceleration that leads to change of direction; rubber bungs are safer than rocks and my favourite demo was bung on string through a pen barrel with a card flag then a weight the other end, thanks to Nuffield. For the actual release, use just a bung on a string, I favoured crochet thread. (Long story there.) Yes, Further to this, the shift in slope at the corners gives an angle as the ant metaphor will show and yes just at the curve it will be a 90 degree angle. This in turn ties to calculus, limits, infinitesimals [duly tamed] and eliminates the perceived contradiction. This brings in how naturally R* is involved in real world situations, even with old fashioned Geometry. KF kairosfocus
SA2, yes the developers should have given such, even a dot to indicate centre. They didn't, and even though from familiarity it is effectively a semicircle (that should have shown intent) the question was asked. I answered it using an old physics data analysis technique: Graph Paper is an instrument. As for walking around with a mirror, those of us who keep a Swiss Army Knife in the pocket do in fact have something close, it will require a little tilting by eye. It happens that a Kitchen knife has a better mirror as it is wider in the blade and pretty flat near the back. KF kairosfocus
Viola Lee: At the endpoints of the semi-circle y = 0, so the tangent is a vertical line. Yup, the slope is undefined but approaches + or - infinity depending on which side we're talking about and which side you approach from. Like we said: all undergraduate calculus stuff. Easy peasy. Tangents to a circle are ALWAYS perpendicular to the radius at the same point. But the circle or semi-circle is NOT the tangent. The shape, as shown, has no right angles. Time for another topic folks. JVL
My comment in 42 was about the more general notion of the direction a point is moving on a curve. The equation of the semicircle shown (assuming the diameter is on the x-axis, centered at zero) is y =-sqr(r^2 - x^2), and the derivative is y' = -x/y. At the endpoints of the semi-circle y = 0, so the tangent is a vertical line. Viola Lee
Viola Lee: If the curve is represented by a function (as a circle is) The top half anyway. Unless we're doing implicit stuff . . . JVL
And, once again Steve, you are right. I wish we knew more about the context in which this problem occurred. I can't imagine what surrounding curriculum made the question appropriate for a seven year old. One of the pictures in the OP says, "There is a difference between the correct answer and the answer at this learning stage." That's poor pedagogy, in my opinion. (Feynman has a good quote about not doing that with students someplace, but I don't know where.) Viola Lee
Kairosfocus “ The mirror technique is quite accurate enough to demonstrate intent:” Not all of us carry mirrors around with us. Wouldn’t be much simpler for the teacher to have stated that the shape was a semicircle? My point is simply that it is impossible to answer the question (true or false) with the information provided. Steve Alten2
Here's a real-world example to illustrate the point the WJM brings up. Imagine swinging a rock on a string. Let go of the string. (Don't do this at home - it's dangerous!) The rock flies off, as we say, at a tangent. If we turn this into a pure math problem, and imagine a point moving around a circle, at any point the point is moving in a particular direction, even though at the "next point" it is moving in a different direction. The direction (angle) it is moving is that which is perpendicular to the radius at that point. That is, the direction of a curve at any point is the direction of the tangent line at that point, and thus any measurement of angles related to the point are in reference to the tangent line. If the curve is represented by a function (as a circle is), then the value of the derivative of the function at that point is the slope of the tangent line. The derivative also represent the rate of change of the function, and is immensely important in both theoretical and practical calculus. Viola Lee
William J Murray: However, if you think about the problem in the abstract, the answer is easy. I agree, using limits the problem is easy. Viola Lee knows. But, the point is that it wasn't meant to be a caculus-level problem. I've seen similar things on state-level exams in the UK. JVL
WJM, at 4 I wrote,
The standard definition of the angle between two curves at their point of intersection is the angle formed by their tangents at that point. The Greeks proved long ago that a radius is perpendicular to the tangent at their point of intersection. Therefore, the circle makes a right angle at the endpoints of the diameter.
Calculus solved the theoretical point you are bringing up quite a few centuries ago. Now the practical issue is a bit different. As I discussed above, it is bad practice to not include important given info in a geometry problem, although we all assume the figure is a semicircle. Also, I went to the website and looked at some of their curriculum materials. In my opinion, they are not very good. The problem in question is not a good question for the public-school age students these materials are designed for. Viola Lee
WJM, the world is an experience not an assumption.I will address the issue in the other thread. The points I am making here lie at the root of calculus. The reality is, our ant pivots to follow the curve C, having arrived at the intersection by travelling along D. That pivot is through an angle. KF kairosfocus
BTW, Zeno's paradoxes only exist if one posits an actual, external world exists. Under MRT, motion, velocity and "space" do not actually exist except as mental experiences - which accounts for the basic answer, "Zeno must be wrong because we actually experience these things." Zeno would only "have to be wrong" if we were talking about an actual, physical external world. The reason we can actually overtake the slower runner, or arrive at a destination, has nothing - ultimately - to do with "decreasing distances" or "motion through space." This is why getting the fundamental question about the actual nature of our existence is kinda important; if you don't get that right, you might wind up with these kinds of paradoxes. William J Murray
So: it is literally impossible, utilizing the intersection point and one next to it on a circle, to acquire a true 90 degree angle from the dissecting line, even if that "next point" is at Planck length distance. William J Murray
KF, If any two succeeding point on a circle are perfectly straight, it's not a circle. It may look that way to the observer of any physical reprsentation; it may feel like it in a physical situation, the circle may be so large it looks like you are turning a right-angle corner between straight, perpendicular roads or hallways. None of that matters; that's mistaking a conceptual problem for a physical one. A curved line cannot intersect with a straight line and produce any angle because angles are made by two intersecting straight lines. You can pick two points of a circle (the one at the point of intersection and one immediately after that in either direction) and us that as your two-point "straight line" reference, but then you are not using the curved line to measure your angle, you're substituting a straight line, mapped by selecting two points on the circle, for the actual line, which is curved. In any event, using the point of intersection and one "next to" the point of intersection to map out your straight line substitute cannot produce an actual right angle, even if it is 89.9999999999999.... degrees. William J Murray
WJM, there is a succession of points in any line, a continuum in fact. Where, as the ant turning to follow the circumference shows, there is an angle involved in a corner [which is a relationship of lines thus of points involved], a vertex of sorts. By going to infinitesimals and/or limits, we can make sense out of that. Similar to, at a certain time and place a car is moving at the rate, 30 miles per hour. Then ponder, doing that around a bend. That shift to a new understanding was the effective answer to Zeno's paradoxes and the like. KF kairosfocus
KF @32, You're talking about a succession of points. There is no succession of points on a circle that is straight, even from the point of the intersection to the next point immediately after that. If those two points were horizontal, the line would continue straight from that. A point does not provide an angle. Since no two points on a circle are straight, it cannot provide an angle which requires two straight lines even if the "line" is only two successive points right next to each other. You can pick a point on the path of a circle and draw a straight line through it to get an angle; if you pick the point right after the intersection to draw your straight line through it and the point at the intersection, it cannot be a 90 degree angle. William J Murray
It seems to me the purpose of the question is to teach the child to recognize the difference between a conceptual problem that can best be solved in the abstract, and a physical problem that is only being represented by a physical image or situation. If you think the problem and the solution lies in the drawing itself, you're likely to get the answer wrong by looking at it for your answer. However, if you think about the problem in the abstract, the answer is easy. William J Murray
WJM, to see the point, imagine you are a microscopic ant crawling along diameter D, then you hit a bend at the intersection with the circumference C, and turn to follow the new line. Through how many degrees do you turn? That turn is a definite angle and in the limit of smallness the angle is . . . 90 degrees. That turn is real, that angle is real, it can be measured or calculated. As you follow the curve you are again turning through a tiny angle at each fresh step or you would go off along the local tangent line. The scale of our ant is of course infinitesimal and the scale of the steps is infinitesimal. KF kairosfocus
I don't get the issue. You can't say the line of the circle is perpendicular at the point of intersection because neither a "point" or a "curve" entails the possibility of an "angle." A line dissecting a circle does not produce "angles" because there is no segment length of a circle that is straight. William J Murray
F/N: As I got thrown a security message in my own thread, here is a mod to 29: >>By looking at infinitesimally altered numbers, we can ponder a tiny all but zero line segment at C_0 and profitably discuss its characteristics, C_0 + dc, to C_0, such that it is straight so angles can be described, where dc is a length shorter than any 1/n for a finite counting number n. In effect, we are exploiting a limit idea that as we get closer and closer to C_0, the curviness gets closer to a straight line. Thus, we have an interpretation of slope at a point using infinitesimally altered real numbers.>> KF kairosfocus
VL & AS2, The mirror technique is quite accurate enough to demonstrate intent: semi-circle, as recognised by eye. Recall, all drawings of geometric objects are strictly representational approximations, this isn't technical drawing. Even with TD, there are tolerances. Then, the stretching to find a side issue is itself telling us something, given the focal issue on the table in the OP: the significance of infinitesimals (in suitably tamed, hyperreals form) in clarifying an issue of fairly common Geometry that in turn elucidates tangents to curves, angles, curves, etc and is of course a gateway to the Calculus. It also brings in a case where the perception of a contradiction draws out relevance of an explanation using infinitesimals in resolving it. Taking reciprocals of strictly finite reals, mileposted by the naturals, we can see that the continuum of course has no defined nearest real coordinates (x1, y1) to the point C_0 along the arc. However we also know that any such real pair will not be on the tangent, which touches a curve at one real-value coordinate point. We can call it the osculating point. (And yes, there is a connexion there to Romances.) By looking at infinitesimally altered numbers, we can ponder a tiny all but zero line segment at C_0 and profitably discuss its characteristics, C_0 + dc, to C_0, such that it is straight so angles can be described, where dc is a length shorter than any 1/n for a finite counting number n. Bizarre, but these have actually been around in some form since classical times, and in a circles context give us an easy way to see why a circle's area is pi * r^2: line up wedges from the top and bottom semicircles to create tiny rectangles and add up to exhaust the half-circumferences, i.e. pi * r long and of course r high. Calculus has been peeking in the window for nearly 2,500 years. This concept leads to area under a curve, volume of a solid of revolution and of course an answer to the mathematically calculated volume of a beer barrel or a wine glass or a light bulb. The first, had traditionally been done by the estimation of the skilled Cooper. We need infinitesimals and therefore also their reciprocals, the transfinite hyperreals. KF kairosfocus
KF, I don't think your mirror idea would resolve the issue if the picture were 0.01° short of 180°. Steve is correct, as I have noted, to point this out. When I taught geometry, at the start of the year I noted a few things we could take as given from the picture, notably that points that looked like they were on a straight line were indeed collinear, because that would have been very tedious to mention in every problem all year long. Other than that, I taught students that you reason from the given info, not just because of how a picture looks. On the other hand, with numerical problems I tried to draw pictures accurately and to scale to give students a real-world experience of whether there answers looked like they made sense. So sometimes problems emphasized working logically from the givens, as in proofs, but other times were more real-world oriented and had considerations beyond the logical reasoning. P.S. to Steve: I always emphasized marking right angle with the "box" symbol when they drew their own pictures to indicate that that was known. Viola Lee
re 25: I would congratulate you for knowing that one must know the given premises in order to draw a logical conclusion! Viola Lee
Following from 25 Maybe using an example from another geometric shape would make my point clearer. We know that the Pythagorus theorem applies to right angle triangles. If I provided a triangle that looks like it might be a right angle triangle, with the lengths of the two shorter sides (A and B) provided, could you answer the following true or false question with complete confidence? “The length of the longest side is the square root of the sum of squares of the other two.” Now, answer the same question if it was either stated that this was a right angle triangle, or a small square was drawn in the corner where the two shorter sides meet. Steve Alten2
Kairosfocus “ The DM Article includes — and I did not excerpt as it was obvious enough — that “The question, accompanied by an image of a semi-circle, asked students: ‘True or false? This shape has two right angles. Explain your answer.’ “” But it was the author of the article who stated that the shape was a semicircle, not the information actually presented to the students. Without providing this information and demanding a true or false answer is like asking you if you have stopped beating your wife and only accepting a yes or no answer. Based on the question asked, and the limited information provided about the shape, would you as the teacher give me marks if I responded that ‘there is insufficient information provided about the shape to determine whether it is a true or false statement?’ Steve Alten2
VL, I am noting that -- and why -- it is effectively settled that it is indeed a semicircle. KF PS: The DM Article includes -- and I did not excerpt as it was obvious enough -- that "The question, accompanied by an image of a semi-circle, asked students: 'True or false? This shape has two right angles. Explain your answer.' " kairosfocus
I only brought up the importance of being very clear when drafting test questions because of an experience I had while marking a university exam. The question was a fill-in-the-blanks. “The Galápagos Islands are or ____ origin and named after ______.” The answers we were looking for were ‘volcanic’ and ‘giant tortoises’. We had a student who answered ‘volcanic’ and ‘1325’. Although it wasn’t the answer we were looking for, it was a correct answer. Steve Alten2
KF, I have no idea why you addressed all that to me??? Yes, we all assumed it was a semicircle. The text in the OP called it a semicircle, but the picture just said "this shape". Steve was right that that original problem in the picture should have specifically stated it was a semicircle. However I think we all agree that we don't know why seven year olds were being asked this question. Viola Lee
VL, behind a planar, front-silvered mirror lies a virtual half universe with images laterally reversed and set back the same distance behind as the objects are in front. This is one legitimate case of a virtual world! One, with rather strict mathematical properties and structures. (Ponder the case of a floating silvered sphere.) As a result of the physics of reflection, if the arc was short of the full circle, the mirror along the secant line would show a lenslike structure, similar to (). Were the secant line beyond the diameter, we would obtain a sort of figure 8, with incomplete balls. Only with a semicircle would we see a rounded, smoothly curved O, a circle; of course, an ellipse would be similar if split symmetrically, a circle is a limiting case of an ellipse, with a parabola another, with the second focal point at infinity. This is what we do see, besides, many people are familiar with the specific shape of a semicircle and the shape in the OP is that. In my native land for example a common food is a patty, which is a folded over pastry circle filled with meat and potatoes to make a working person's meal. The geometry half circle protractor is another common case, though of course that goes a bit beyond the circle. Though it is not stated and the centre is not indicated, the figure is and is intended to be a semicircle. KF kairosfocus
Also, Kf's picture above shows how you transform a secant line (cutting the curve at two points) into a tangent line by moving the second point (on the right in his picture) closer and closer to the fixed point, so that the secant becomes the tangent. The moving point can come from either the right or the left, and if the curve is continuous and differentiable, you will get the same result from either direction.. However, if you're at at the endpoint of a curve, as we are in this situation, you can come form whatever direction is appropriate, and the result you get is still the tangent line. Maybe that helps. Viola Lee
You're right, Steve. The problem didn't say specifically that it was a semicircle. I written lots of geometry problems, and one convention is to label the midpoint of the diameter, often with O, to clarify that it went through the center of the circle so that the arcs were semicircles. Geometry problems should state the givens, and this one didn't. Viola Lee
If the definition of a circle is x^2 + y^2 = R^2, but the definition of a right angle, per Pythagorus is that the square root of both sides of the above equation are equal, then even with infinitesimals, R^2 > (x^2 + y^2)^1/2 , and the result is an inequality. But I imagine analysis can find a way around such restrictions. PaV
SA2, the physical confirmation rests on the laws of reflection, i.e. incident, reflected ray and normal to plane of reflection are in a common plane and incident and reflected angles are equal. If your monitor is a flat screen, you can do it readily as I did. KF kairosfocus
I will definitely defer to you and Viola Lee on this. Steve Alten2
SA2, it is both implied and visually recognisable. KF PS: The mirror test will show it instantly. PPS: Using a polished kitchen knife blade, a lab not being handy, it is confirmed to be a semicircle. kairosfocus
Kairosfocus, thank you for the visual. However, I hear terms like circumference, diameter and radius being used to describe the solution to the question. But that involves reading more into the question than was provided. You are assuming that the shape displayed is a semi-circle, but that was not stated. Steve Alten2
SA2, the easiest answer is a physical one. Take a small, flat, preferably front silvered mirror though practically a back silvered one will do. Pivot it at the point where the diameter D hits the circumference C. Move it on the page in a rocking motion until the reflection goes without a break in the mirror image. If your hands are shaky, stretch out a little finger and use it as a damped prop against the page to stabilise. Trace the line of the mirror. Oddly, it also works for a reasonably smooth curve: do the normal to the curve then the perp to the normal to get the tangent. It is easier, quicker and more accurate than most things, exploiting our ability to detect straightness by eye. KF PS: For the Geometric construction, produce the line of the diameter beyond the circumference by several inches. Use a compass pivoted on the intersection of D and C, to spin a semicircle standing on the produced diameter, then on its intersections with D, cut arcs on the new semicircle at the 60 degree points. At these cuts, make arcs that will intersect on the tangent line. Set a straight edge from the point where the original C and D intersect to the tangent line intersection and draw the line. To set a compass at a point, lean it over and move it so the tip is where you want the centre, then carefully tilt back up. Spring Bow compasses are the most accurate. PPS: Also, just use a set square sitting on the diameter. kairosfocus
Viola Lee, forgive me for possibly being dense. I admit that any radius that intersects a tangent of a circle at the same point will do so at right angles. But the shape shown in the question is not a circle. It may be a semi-circle, but that has not been stated. For any point along an arc there can only be a single slope for a tangent that touches that point, i.e., it can't pivot on that point without crossing the plane of the arc (Forgive me if I am misusing terms. I wish I could post diagrams to illustrate what I am trying to say). However, on either side of the shape there is a point which is the distal end of the arc and the distal end of the straight line. I can draw multiple lines, I don't think you can call them tangents, through this point that do not cross the plane of the shape. I hope that I am making myself clear on this. Maybe I should number my sentences. :) Steve Alten2
Steve, you just draw it perpendicular to the radius at that point. The fact that the semicircle doesn't continue on doesn't affect the tangent line. Viola Lee
Kairosfocus, yes, I understand this. And for a circle, any tangent will be at a right angle to the centre of the circle. You can't have any tangent to a circle that isn't at a right angle to the centre. But how do you draw a tangent at the point where the semi-circle changes from a curve to a straight line? Steve Alten2
Viola Lee @4, Nicely stated. One could also use tangents on a full circle with a diameter through it. This sort of thing is common in schools and on tests. The SAT was once famous for a question involving three integers and asking what the forth one should be. One bright student was able to challenge the question by demonstrating that one of the "wrong" answers could indeed be right. Here's an example: What's the next element in the series, 1, 2, 3, . . . ? A. 4 (add one to the previous element) B. 5 (add the next number in the Fibonacci series to the previous element) C. 10 (base 4) D. None of these (can the test taker find any real number in the series using any other method?) -Q Querius
SA2, Actually, slice a circle in half and toss the half you don't want. Where, our illustrations are just that, visualisations of something that, strictly is abstract. We cannot draw a perfect circle, by ink or by PC, What we call lines are actually stripes or bands, strictly a line has no breadth. What we mark as a point, has no breadth or width, the dot is just an idea of the actual point. Also, a tangent touches the line at just one point. Things like this are why the algebraic representation is more exact. KF PS: This kind of stuff suggests how Plato was led to think of a world of forms. kairosfocus
I agree that this is not a question for a seven year old, and probably not even for a grade seven. And, depending on the response to my comment below, this may not even be a question for a 62 year old. I am not a mathematician but given the question posed in the screen-shot in the OP, "This shape as two right-angles", I would have to say that it can't be answered without more information. It certainly looks like a semi-circle, but this isn't stated. It could just as easily be slightly less or more than a semi-circle, in which case there could not be right-angles. But let's assume that it is a semi-circle. It is obvious that a line drawn between the intersection point of a tangent to a circle and the circle's centre will be a right angle. But how do you draw a tangent at the exact point where the curved part of the semi-circle stops being a continuous curve (takes a sharp left or right?)? Steve Alten2
To polistra: If there were just a 180° arc by itself, without the diameter, there wouldn't be any angles at all. Viola Lee
VL, this came from an introduction to the concept of angles. Likely, intended as a counter-example. KF kairosfocus
This is not a question for seven year olds. But the standard definition of the angle between two curves at their point of intersection is the angle formed by their tangents at that point. The Greeks proved long ago that a radius is perpendicular to the tangent at their point of intersection. Therefore, the circle makes a right angle at the endpoints of the diameter. QED. Viola Lee
P, a key issue here is what does it mean to be curved. KF kairosfocus
Even without using calculus the picture itself is perfectly clear. Two right angles. If they were asking about an abstract semicircle without picturing it, or a 180 degree arc by itself, then it wouldn't be clear. polistra
Semi-circles and right angle dilemmas . . . implications and issues for a math puzzle for seven year old students. Yes, infinitesimals at work. kairosfocus

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