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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?

Thanks!

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What do you mean by D (D /ne A)? –  leo Sep 27 '12 at 2:07
    
He means $D\ne A$, that is, D\ne A –  Pedro Tamaroff Sep 27 '12 at 2:33
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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? –  user31029 Sep 27 '12 at 3:32
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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. –  Willie Wong Sep 27 '12 at 10:40
    
This forum is suited better for olympiad problems: artofproblemsolving.com/Forum/portal.php?ml=1 –  Dominik Dec 10 '12 at 20:53
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2 Answers 2

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

description

  • 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 $CD=3521$
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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: $$ CE = 4024 * \sin(\angle CAB) \approx 1948 $$ $$ CD = 2 * CE \approx 3896$$

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