Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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

2 Answers 2

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 $DE^2=5\sqrt{3}/2$.

share|improve this answer

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 $DE^2=AD\cdot DB=5\sqrt{3}/2$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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