Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed.

I am preparing for competition math and so I found this problem I was wondering if someone could post a solution to it because I can't solve it. This is from a book in the Art of Problem Solving series I am sure some of you have heard of it.

We let ABC be a triangle with sides: AB=4024, AC=4024 and BC=2012 . We reflect the line AC over line AB to meet the circumcircle of triangle ABC at point D (D $\ne$ A). How do we find the length of segment CD?


  • $\begingroup$ What do you mean by D (D /ne A)? $\endgroup$ – leo Sep 27 '12 at 2:07
  • $\begingroup$ He means $D\ne A$, that is, D\ne A $\endgroup$ – Pedro Tamaroff Sep 27 '12 at 2:33
  • 11
    $\begingroup$ Mods, please delete this.This question is from an ongoing contest called online math open. onlinemathopen.netne.net/sites/default/files/OMOFall12.pdf See problem no. 16.Do not post solutions here . Is there a way this IP can be reported to the OMO organizers? $\endgroup$ – user31029 Sep 27 '12 at 3:32
  • 5
    $\begingroup$ Actually, here's a better link: meta.math.stackexchange.com/questions/4004/… The community consensus seems to be that if you see something, tell the contest organisers about it. A "no contest question" policy would be very difficult to enforce, and as I mentioned above we cannot give out the IP addresses. $\endgroup$ – Willie Wong Sep 27 '12 at 10:40
  • $\begingroup$ This forum is suited better for olympiad problems: artofproblemsolving.com/Forum/portal.php?ml=1 $\endgroup$ – Dominik Dec 10 '12 at 20:53

From the link, the contest should be over by now, so a solution definitely won't hurt.


  • Draw the triangle $\triangle ABC$, the reflected line $AC'$ and the circumcircle. Label the intersection of $AC'$ with the circle $D$
  • Obviously $CC$ and $AB$ form a right angle at the intersection since $AB$ is the center of reflection. Draw the segments $CC'$, $BC'$, $BD$ and $CD$.
  • Label the intersection of $CD$ and $AB$ point $E$.
  • By congruency $SAS$, observe that $\triangle ABC \equiv \triangle ABC'$
  • I'm treating a general case where the triangle has lengths $2x$, $2x$, and $x$. So we have $$\begin{align}&AB=2x\\ &AC'=AC=2x\\&BC'=BC=x \\ & \angle ABC=\angle ABC'=\angle ACB=\angle AC'B \\& \angle BAC=\angle BAC' \end{align} $$
  • By inscribed angle theorem we have the following

    1. $\angle BDC =\angle BAC = \angle BAC'=\angle BCD$

      $ \implies \triangle BDC$ is isosceles and $BD=BC=BC'=x \implies \triangle BCC'$ and $\triangle BC'D$ are isosceles thus $\angle BC'D=\angle BDC' =\angle BC'A=\angle ABC$

    2. $\angle CBA =\angle CDA = \angle BC'D$ observe that points $D$ and $C$ are on the same line thus $\angle BC'D = \angle CDA \iff CD \parallel BC' $
  • Since $ED \parallel BC'$, then $\triangle ADE$ is not only isosceles but similar to triangle $ \triangle ABC' \equiv \triangle ABC$
  • $\angle CEB =\angle AED =\angle ADE =\angle CBE \implies \triangle CBE$ is isosceles with $CE=CB=x$
  • At this point, it is sufficient to find $DE$ in order to find $CD$.
  • From similar triangles (of $\triangle BEC$ and $\triangle BCA$) we have $BE=C'D= \cfrac x2$ and $AE=AD=2x-\cfrac x2$ and by similar triangle again $\cfrac {DE}{C'D}=\cfrac {AE}{AB} \implies \cfrac{DE}{x}=\cfrac {2x-\cfrac x2}{2x}\implies DE= x-\cfrac x4$
  • Finally $CD=CE+ED=x+x-\cfrac x4 =2x-\cfrac x4$
  • In the case where $x=2012$, we have $CD=3521$

This can be solved using trigonometry. First, find the angle CAB: $$ \angle CAB = 2 * \sin^{-1}(1006/4024) \approx 29^{\circ} $$ Now, label the intersection between CD and AB point E. Using more trig: $$ CE = 4024 * \sin(\angle CAB) \approx 1948 $$ $$ CD = 2 * CE \approx 3896$$

  • $\begingroup$ it is $3521$ not $3896$ $\endgroup$ – user31280 Dec 8 '12 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.