$ABC$ and $ODE$ are equilateral triangle with $BC || DE$. If $O$ is the center of the circle, then find the ratio $AQ:QC$

So, my thought on this is that, since we are not given the length, let $AO = x$

Then, the perpendicular $OZ$ falling on line segment $BC$ is $\frac{2}{3}x$ (I am not sure)

So, using $$\text{Area using Heron's formula} = \text{Normal area}$$

$$\Rightarrow \sqrt{\frac{3x}{2} * (3 * (\frac{3x}{2} - x))} = \frac{1}{2}*(\frac{2x}{3} + x) * x$$

$$\Rightarrow \frac{9x^2}{4} = \frac{2x^2 + 3x^2}{6}$$

$$\Rightarrow \frac{17x^2}{12} = 0$$

which gives $x = 0$ (not possible!)

Can anyone please tell what should be the approach to this problem.

Thank you.

Reference Diagram (courtesy Alex R):


  • $\begingroup$ I think a better picture would help you: since your triangles are equilateral, they all have 60° angles, and many other nice properties. By the way, since everything is scalable, you may do all your computations with the unknown $x$ or simply assume that $x=1$. In the end, it should cancel out anyway. $\endgroup$ – zozoens Jan 30 '14 at 15:43
  • $\begingroup$ @zozoens Yes, in one of my old calculations, I eventually cancelled out x and got $something = 1$ ! So, I tried again and again and got this (which again is wrong). And I will try to give a better diagram. Two minutes. $\endgroup$ – Gaurang Tandon Jan 30 '14 at 15:45
  • $\begingroup$ @zozoens Added image. $\endgroup$ – Gaurang Tandon Jan 30 '14 at 15:50
  • $\begingroup$ This is how I found a result that seems correct: using properties of the triangle $AQO$, compute $AQ$ (depending on $x$). Then compute $AC$ using properties of equilateral triangle $ABC$. Then the quotient is directly computable. $\endgroup$ – zozoens Jan 30 '14 at 15:53
  • $\begingroup$ @zozoens Thanks, but I am having trouble figuring out the properties of the AQO (it's a scalene triangle anyway). Could you please post an answer with some more detail (but let me figure the answer out). Thank you. $\endgroup$ – Gaurang Tandon Jan 30 '14 at 15:55

This sketch (made with GeoGebra) should do better than a hand-drawn. All Specs are met (both triangles are equilateral, $BC\parallel DE$ and the circle is in fact a circle. enter image description here

From this we can use the intercept theorem for $PQ\parallel BC$ and thus $AQ:QC = AR:RS$ But since $AR=OR$ by construction, $RO = OS = AR$ and therefore $$AR:RS = 1:2=AQ:QC$$

  • $\begingroup$ Very nice drawing . Edited my question to include this image. I will check out GeoGebra soon. $\endgroup$ – Gaurang Tandon Jan 30 '14 at 15:57
  • $\begingroup$ Nice. But this isn't an answer. $\endgroup$ – David Mitra Jan 30 '14 at 15:59
  • $\begingroup$ @GaurangTandon considering your recent questions, I can definately recommend it to you ;) Thanks for using $\endgroup$ – AlexR Jan 30 '14 at 16:02
  • $\begingroup$ @DavidMitra yeah, but too big for a post and too massive for an edit on my own. I'll leave it here since it could be considered some sort of hint^^ $\endgroup$ – AlexR Jan 30 '14 at 16:04
  • $\begingroup$ @GaurangTandon Take a look at my new picture, this should answer everything... $\endgroup$ – AlexR Jan 30 '14 at 16:30

Calling $Q$ the intersection point of the lines $(AC)$ and $(OE)$, one can check that $AQO$ is isocele in $Q$, since both the angles $\widehat{Q A O}$ and $\widehat{Q O A}$ are $\pi/6$. From this one can compute the length of its sides depending on the length of its basis (do some trigonometry in one of its half-triangles).

Computing $AO$ from $AC$ is a classical exercise of equilateral triangle properties.

So we get both $AC$ and $AQ$ from $AO$, and their quotient (almost) gives the desired result.

  • $\begingroup$ I know this is really basic a question, but how the angles $QAO$ and $QOA$ are $\frac{\pi}{6}$ ? $\endgroup$ – Gaurang Tandon Jan 30 '14 at 16:10
  • $\begingroup$ @GaurangTandon $AO$ halves the angles $\angle CAB$ and $\angle EOD$, which are both $60^\circ$ due to the triangles being equilateral. $\endgroup$ – AlexR Jan 30 '14 at 16:18
  • $\begingroup$ @zozoens This sounds good, but shouldn't $QAO$ and $QOA$ then be $30\degree$ each, instead of $\frac{\pi}{6}$, since $AO$ halves ∠CAB and ∠EOD. $\endgroup$ – Gaurang Tandon Jan 30 '14 at 16:32
  • $\begingroup$ $30°=\pi/6$ rad $\endgroup$ – zozoens Jan 30 '14 at 16:59

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