You are given a circle in the plane whose center is marked. Is it possible to construct a pair of perpendicular diameters of this circle using only a straightedge?

It is possible via the Poncelet–Steiner theorem. By reviewing the proof, you can figure out how to do it. The relevant sections are Prelim. 1a. and Prelim 2. Construct a diameter of the circle. Make a line parallel to it that is not a diameter and that intersects the circle at two points U and V (Prelim 1a). Construct the diameter UU'. Connect U' and V. U'V is perpendicular to the original diameter. Make a line parallel to it that crosses the origin (Prelim 1a) and you're done. I'm not sure if there's a simpler solution.

@Tim Vermeulen
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I can indeed. I don't know -- perhaps by analyzing all the details carefully, one could discover that this method is essentially equivalent to the more general approach that Colby proposed. However, it should at least be easier to remember. I should also give credit to my friend Apoorva K. for sharing this problem and explaining the solution to me.

Step 1. Draw an arbitrary diameter of your circle. Then show that it is enough to construct one chord of the circle that is perpendicular to this diameter. (Let me know if you want additional details for this step; it is not hard to figure out what to do once you have a chord that is perpendicular to your diameter, but it's kind of a pain to write out all the steps.)

Step 2. The fun part is getting a chord perpendicular to your diameter. Let's say that your diameter is AB. Choose two arbitrary points C and D on your circle, so that they are on the same side of the line AB and such that the order in which the four points appear on the circle is this: A, C, D, B. Draw the lines AC and BD. Call E their point of intersection. Now consider the triangle \(\triangle\)AEB.

Note that the angle \(\angle\)ADB is a right angle (because AB is a diameter of your circle and D lies on the circle). So the line AD is an altitude of the triangle \(\triangle\)AEB. Similarly, the line BC is an altitude of the triangle \(\triangle\)AEB. Let F be the intersection point of AD and BC. Then, by a famous theorem, the line EF is also an altitude of the triangle \(\triangle\)AEB -- and there you have your chord that is perpendicular to the diameter AB.

A simpler solution after getting a parallel chord:

Call the diameter \(AB\) and the parallel chord \(UV\) such that \(A,U,V,B\) appear in that order along an arc. Draw \(AV\) and \(BU\) and call the intersection \(M\). Call the center \(O\); draw the line \(MO\). This is a diameter perpendicular to \(AB\).

My proof utilizes the fact that \(AUVB\) is isosceles and hence the entire thing is symmetric over \(MO\). We can relabel \(A \to B, B \to A, U \to V, V \to U\) and then reflect the entire thing over a line perpendicular to \(AB\). \(MO\) will have a different slope except if it's perpendicular or parallel to \(AB\). The latter case is impossible as \(M\) won't be on \(AB\) (\(AV\) crosses \(AB\) at \(A\), \(BU\) crosses \(AB\) at \(B\), and \(A \neq B\), so the intersection \(M\) cannot be on \(AB\)), so \(MO\) must be perpendicular to \(AB\).

I agree. In fact, your observation can also be used in Step 1 of the solution that I posted, because once you have a perpendicular chord, you can immediately construct a parallel chord (if UW is a perpendicular chord, O is the center of the circle and V is the second intersection point of WO with the circle, then UV is a parallel chord).

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## Comments

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TopNewestIt is possible via the Poncelet–Steiner theorem. By reviewing the proof, you can figure out how to do it. The relevant sections are Prelim. 1a. and Prelim 2. Construct a diameter of the circle. Make a line parallel to it that is not a diameter and that intersects the circle at two points U and V (Prelim 1a). Construct the diameter UU'. Connect U' and V. U'V is perpendicular to the original diameter. Make a line parallel to it that crosses the origin (Prelim 1a) and you're done. I'm not sure if there's a simpler solution.

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That's a powerful result! I didn't know this approach before -- the solution I've seen was specific to this problem.

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Could you share it?

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Step 1. Draw an arbitrary diameter of your circle. Then show that it is enough to construct one chord of the circle that is perpendicular to this diameter. (Let me know if you want additional details for this step; it is not hard to figure out what to do once you have a chord that is perpendicular to your diameter, but it's kind of a pain to write out all the steps.)

Step 2. The fun part is getting a chord perpendicular to your diameter. Let's say that your diameter is

AB. Choose two arbitrary pointsCandDon your circle, so that they are on the same side of the lineABand such that the order in which the four points appear on the circle is this:A,C,D,B. Draw the linesACandBD. CallEtheir point of intersection. Now consider the triangle \(\triangle\)AEB.Note that the angle \(\angle\)

ADBis a right angle (becauseABis a diameter of your circle andDlies on the circle). So the lineADis an altitude of the triangle \(\triangle\)AEB. Similarly, the lineBCis an altitude of the triangle \(\triangle\)AEB. LetFbe the intersection point ofADandBC. Then, by a famous theorem, the lineEFis also an altitude of the triangle \(\triangle\)AEB-- and there you have your chord that is perpendicular to the diameterAB.Log in to reply

A simpler solution after getting a parallel chord:

Call the diameter \(AB\) and the parallel chord \(UV\) such that \(A,U,V,B\) appear in that order along an arc. Draw \(AV\) and \(BU\) and call the intersection \(M\). Call the center \(O\); draw the line \(MO\). This is a diameter perpendicular to \(AB\).

My proof utilizes the fact that \(AUVB\) is isosceles and hence the entire thing is symmetric over \(MO\). We can relabel \(A \to B, B \to A, U \to V, V \to U\) and then reflect the entire thing over a line perpendicular to \(AB\). \(MO\) will have a different slope except if it's perpendicular or parallel to \(AB\). The latter case is impossible as \(M\) won't be on \(AB\) (\(AV\) crosses \(AB\) at \(A\), \(BU\) crosses \(AB\) at \(B\), and \(A \neq B\), so the intersection \(M\) cannot be on \(AB\)), so \(MO\) must be perpendicular to \(AB\).

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I agree. In fact, your observation can also be used in Step 1 of the solution that I posted, because once you have a perpendicular chord, you can immediately construct a parallel chord (if

UWis a perpendicular chord,Ois the center of the circle andVis the second intersection point ofWOwith the circle, thenUVis a parallel chord).Log in to reply