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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

Comments
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
March 5, 2021
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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
March 5, 2021
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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
March 5, 2021
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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. KFkairosfocus
March 5, 2021
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I will definitely defer to you and Viola Lee on this.Steve Alten2
March 5, 2021
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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
March 5, 2021
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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
March 5, 2021
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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
March 5, 2021
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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
March 5, 2021
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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
March 5, 2021
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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
March 5, 2021
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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?) -QQuerius
March 5, 2021
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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
March 5, 2021
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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
March 5, 2021
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To polistra: If there were just a 180° arc by itself, without the diameter, there wouldn't be any angles at all.Viola Lee
March 5, 2021
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VL, this came from an introduction to the concept of angles. Likely, intended as a counter-example. KFkairosfocus
March 5, 2021
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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
March 5, 2021
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P, a key issue here is what does it mean to be curved. KFkairosfocus
March 5, 2021
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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
March 5, 2021
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Semi-circles and right angle dilemmas . . . implications and issues for a math puzzle for seven year old students. Yes, infinitesimals at work.kairosfocus
March 5, 2021
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