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In traingle ABC, Angle A=45 degrees, Angle B is 60 degrees, and AC= radical 15. D is also a point on AB so that AB is perpendicular to CD. The circle with diameter AB intersects CD at point E. Compute (DE)^2.

So this is what I have done so far, since I just joined today and I never knew I had to post what work I done so far. Since we know A and B, we know that C is 75 degrees. Since AB is perpendicular to DC we know that triangle ADC is isosceles and traingle DBC is a 30, 60, 90 triangle. I could find the lengths of the sides of triangle ADC and triangle ADC but how would I find DE?

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Aren't we missing some info? The problem description doesn't seem to pin down point $B$ (though there are restrictions). Yet, the longer $AB$ is, the longer $DE$ is, so that we have no unique solution. If $B$ and $D$ coincide, then $E$ coincides with them, making $DE = 0$. If $B$ is even a hair further away from $A$, then $DE$ is non-zero. The other extreme is when the circle is large enough so that $E$ matches $C$, in which case $\angle ACB$ is a right angle (why?) and $AB =$ ___. In any (non-zero) case, $\angle AEB$ is a right angle (same "why?"), so that $AD/DE=ED/DB$ (different "why?"). – Blue Oct 27 '11 at 2:47
BTW, I agree with what has been noted in a comment to another of your (non-)questions: unmotivated contest problems aren't appropriate for this site (though I do appreciate that you've made an effort to note the source). I chose to "answer" above because I found the problem, as posed, unsolvable, and found that to be interesting. :) – Blue Oct 27 '11 at 3:00
Believe me I have tried to answer this problem. I have recieved the answer after the competition was completed. But it still seems unsolvable to me at this point >_< – Tayyeb Din Oct 27 '11 at 3:10
Ah ... You've edited to include the measure of angle $B$! Okay, then: Now, you can get the length of $DB$. (Use the Law of Sines to get $AB$ and go from there, or else use the geometry of the 45-45 triangle $ADC$ and 30-60 triangle $BDC$. (The latter is probably what the problem intends.)) Then, you can use the proportion at the end of my first comment. (The proportion follows from that fact that $\angle AEB$ is a right angle; as is any angle (as they say) "inscribed in a semi-circle".) – Blue Oct 27 '11 at 3:54
Thank you! I am not used to using proportions. But I really appreciate the help. – Tayyeb Din Oct 27 '11 at 4:16

In order to keep the numbers, and more importantly, the typing, simple, define $k$ by $\sqrt{15}=2k\sqrt{6}$. We do some side-chasing, using properties of so-called "special" triangles.

Since $\angle CAD=\angle DCA=45^\circ$, we have $AD=CD=2k\sqrt{3}$. And since $\angle ABC=60^\circ$, the tangent of this angle is $\sqrt{3}$, and therefore $DB=2k$.

Thus $AB=2k(\sqrt{3}+1)$, and therefore the circle has radius $k(\sqrt{3}+1)$. Also, if $O$ is the center of the circle, then $OD=AD-AO=2k\sqrt{3}-k(\sqrt{3}+1)=k(\sqrt{3}-1)$.

By the Pythagorean Theorem for $\triangle ODE$, we have $$DE^2=OE^2-OD^2=k^2(\sqrt{3}+1)^2-k^2(\sqrt{3}-1)^2=4\sqrt{3}k^2.$$
But $k^2=5/8$, so $DE^2=5\sqrt{3}/2$.

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You found $AD=\sqrt{15}/\sqrt{2}$ and $DB=(\sqrt{15}/\sqrt{2})/\sqrt{3}$. Now the triangle $AEB$ is rectangular, therefore by the "altitude theorem" one has $DE^2=AD\cdot DB=5\sqrt{3}/2$.

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