An isosceles triangle with a unit length base is sliding on two lines which make an angle of $60^\circ$ between them. The third vertex traces a circle centered at the intersection of the two lines. What is the altitude of the triangle, and what is the radius of the circle ?

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What I have tried:

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I found the coordinates of points $X$ and $Y$ as shown above on the two lines in terms of the angle $\theta$, then found the coordinates of the tip of the triangle (the third vertex) as a function of $\theta$, and finally found the altitude $h$ that results in the distance of this tip from the origin being constant. And that constant is the radius of the circle.

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    $\begingroup$ What have you tried? Where does this exercise come from? Is it a contest question? $\endgroup$ Feb 28, 2022 at 10:08
  • $\begingroup$ No. It's not a contest question. It's something that I came up with. I solved the problem and would like to see how others solve it. $\endgroup$
    – Hosam H
    Feb 28, 2022 at 10:11
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    $\begingroup$ But it would be very interesting to see your solution first (yes, I am always sceptical here, sorry). You can add that you are looking for a different solution. $\endgroup$ Feb 28, 2022 at 10:12
  • $\begingroup$ Please show us your solution so as to avoid unnecessary duplication of effort. $\endgroup$ Feb 28, 2022 at 10:41
  • $\begingroup$ That really cool. Does it only work for $60$ degrees? $\endgroup$ Feb 28, 2022 at 12:04

3 Answers 3


A very fast synthetic approach

As in your notation, let $\phi$ and $\pi -\phi$ be the angles between the straight lines intersecting in $O$.

We want to construct an isosceles triangle with sides $BC \cong AC$, and vertex $C$ on the circle centered in $O$ having radius $\overline{BC}$.

It is sufficient then to analyse the two cases shown below.

Case 1

Let $\measuredangle AOC = \alpha$. Since $\triangle AOC$ is isosceles we have $\measuredangle ACO = \pi -2\alpha$. Similarly, $\triangle BOC$ is isosceles, and $\measuredangle BOC = \pi - \phi -\alpha$, so that $\measuredangle BCO = \pi - 2(\pi-\phi-\alpha) = 2\phi + 2\alpha - \pi$. Therefore $\measuredangle ACB = 2\phi$.

Case 2

Let now $\measuredangle A'OC' = \beta$. Since $\triangle A'OC'$ is isosceles we have $\measuredangle A'C'O = \pi -2\beta$. Similarly, $\triangle B'OC'$ is isosceles, and $\measuredangle B'OC' = \phi -\beta$, so that $\measuredangle B'C'O = \pi - 2\phi +2 \beta$. Thus $\measuredangle A'C'B' = 2\pi - (\pi -2\beta) - (\pi - 2\phi +2\beta) = 2\phi$.

Your thesis follows immediately, since the previous results do not depend on the choice of either $\alpha$ or $\beta$.

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  • $\begingroup$ Thanks for the answer. I have one question, how did you make the drawing, is it with GeoGebra, or some other software ? $\endgroup$
    – Hosam H
    Mar 1, 2022 at 14:56
  • $\begingroup$ @LexiBelleFan You're very welcome. It's Geogebra, indeed. $\endgroup$
    – dfnu
    Mar 1, 2022 at 15:15
  • $\begingroup$ @dfnu I have given a solution using complex numbers geometry. $\endgroup$
    – Jean Marie
    Mar 7, 2022 at 14:30

The solution by @dfnu is indeed simple.

But something even simpler can be given (avoiding in particular two cases).

Let $\varphi$ be the angle between the two "guiding" axes $A_1$ (horizontal) and $A_2$ intersecting in $O$, where $\varphi$ can be any value from $0$ to $\pi/2$, generalizing the case $\varphi=\pi/3$ given in the question.

Let us work backwards. Let us consider a circle with center $O$ ; we can assume WLOG that its radius is $1$.

With complex notation $e^{i\alpha}$ for the current point $C$ of the circle, let us consider the triangle with vertices $A,B,C$ defined in this way:

$$\begin{cases}A&=&e^{i\alpha}+e^{-i\alpha}&=&2\cos(\alpha),\\ B&=&e^{i\alpha}+e^{-i\alpha}e^{2i\varphi}&=&2 \cos(\alpha-\varphi)e^{i\varphi},\\ C&=&e^{i\alpha}&&\end{cases}\tag{1}$$

(use Euler formulas for converting between the two forms for $A$ and $B$).

$A$ belongs clearly to the horizontal axis $A_1$ and $B$ to axis $A_2$. Furthermore $CA=CB$ because $|A-C|=|B-C| \ \iff \ |e^{-i \alpha}|=|e^{i (-\alpha+2 \varphi)}|=1$ proving that $ABC$ is an isosceles triangle.

Some particular cases are worth to be considered: $\varphi=\pi/6,0,\pi/2$.

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Fig. 1: $\varphi=\pi/6$ gives equilateral triangles.

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Fig. 2: The degenerated case $\varphi=\pi/2$ (orthogonal axes $A_1$ and $A_2$) with flat triangles as well. The envelope of these lines is an astroid. This is the famous problem of the ladder falling down ; see the nice animation here.

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Fig. 3: Another degenerated case $\varphi=0$ where axes $A_1$ and $A_2$ are superimposed gives flat triangles. One can observe that it is in fact a part of the previous figure (the envelope is the middle part of the astroid).

Remark: I discovered (thanks to @brainjam) that this is known as the van Schooten's mechanism. A connected presentation can be found as well here.


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Let one line be horizontal and the other making an angle of $\phi$ with it.

I'll take the origin of the coordinate system at their intersection. Placing the triangle at a general orientation with its base making an angle of $\theta$ with the horizontal as shown above, we can write the following equation from the law of sines

$ \dfrac{y}{\sin \theta } = \dfrac{1}{\sin \phi} $

Hence, $ y = \dfrac{\sin \theta}{\sin \phi} $

Since the unit vector pointing from the origin to $Y$ is $(-\cos \phi, -\sin \phi) $, then point $Y$ has the following coordinates

$Y = (- \sin \theta \cot \phi, - \sin \theta ) $

Now, the unit vector along $\vec{YX}$ is $(-cos \theta, \sin \theta)$, therefore,

$X = Y + (1) (-\cos \theta, \sin \theta) = (- \sin \theta \cot \phi - \cos \theta, 0 ) $

The midpoint of the base has coordinates

$ M = \frac{1}{2} (X + Y) = Y + \frac{1}{2} ( - \cos \theta, \sin \theta ) $


$M = (- \sin \theta \cot \phi - \frac{1}{2}\cos \theta, - \frac{1}{2} \sin \theta ) $

Finally the tip of the triangle $T$ is a distance $h$ in the direction $(\sin \theta, \cos \theta) $, thus its coordinates are

$T = ( \sin \theta (h - \cot \phi) - \frac{1}{2} \cos \theta, -\frac{1}{2} \sin \theta + h \cos \theta ) $

If $T$ traces a circle centered at the origin, then the sum of squares of its $x$ and $y$ coordinates must be constant, hence

$ T_x^2 + T_y^2 = \sin^2 \theta ( (h - \cot \phi)^2 + \frac{1}{4} ) + \cos^2 \theta ( \frac{1}{4} + h^2 ) - \sin \theta \cos \theta ( 2 h - \cot \phi ) $

For this to be constant, we must have

$ (h - \cot \phi)^2 = h^2 + \frac{1}{4} $


$ 2 h - \cot \phi = 0 $

We can see that the two equations are identical and have the solution

$ h = \frac{1}{2} \cot \phi $

And the radius of the circle is $\sqrt{ \frac{1}{4} + h^2 } = \dfrac{1}{2 \sin \phi } $

For $\phi = \dfrac{\pi}{3} $, we get

$ h = \dfrac{1}{2 \sqrt{3}} $ and $R = \dfrac{1}{\sqrt{3} } $

  • $\begingroup$ Have you seen my proposal ? Complex numbers geometry often provides interesting approaches when angles are stronly present... $\endgroup$
    – Jean Marie
    Mar 7, 2022 at 18:25
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    $\begingroup$ It is quite inventive [+1]. (I've already upvoted it). $\endgroup$
    – Hosam H
    Mar 7, 2022 at 20:16
  • $\begingroup$ Have you seen the connected issue that I mention at the bottom ? $\endgroup$
    – Jean Marie
    Mar 7, 2022 at 20:19
  • $\begingroup$ Yes. I am aware of this connected issue, and have implemented the same exact thing about two or three years ago, just like this problem here, it dates back to two or three years ago, that's when I discovered it on my own, just like I discovered Rodrigues rotation formula independently, but that was about 8 or 9 years ago. $\endgroup$
    – Hosam H
    Mar 7, 2022 at 20:26

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