an interesting highschool geometry question

Posted: April 1, 2013 in Mathematics

During the long easter weekend, I finally have time to check out the web and forums and I found this nice little geometry problem on triangles.  At first glance the problem looks difficult, however it is not too bad after figuring out some of the angles and realizing couple isosceles triangles in it.

Solution

First of all let’s label the points A, B, C, D, E, and F in the diagram.

Obviously, triangle ABC, triangle BCD, and triangle ABF are isosceles triangles.  Let

$AB=AC=BF=k, \quad BD=BC=a$

and let

$CD=b, \quad AD=c$

In isosceles triangle ABC, we have

$\dfrac{a}{2k}=\cos{48}$

$a=2k\cos{48}$

In isosceles triangle BCD, we have

$\dfrac{b}{2a}=\cos{72}$

$b=2(2k\cos 48)\cos 72=4k\cos 72 \cos 48$

Then apply the Product-to-Sum identity

$\cos{A}\cos{B}=\dfrac{\cos(A+B)+\cos(A-B)}{2}$

We have

$b=2k(\cos 120 + \cos 24)=(2\cos24-1)k$

Now apply the cosine law on triangle ACD,

$c^2=k^2+b^2-2kb\cos 24$

$c^2=k^2+(2\cos24-1)^2k^2-2k((2\cos24-1)k)\cos24$

$c^2=k^2(1+4\cos^2{24}-4\cos{24}+1-4\cos^2{24}+2\cos{24})$

$c^2=k^2(2-2\cos24)$

And then finally apply sine law in triangle ABD,

$\dfrac{\sin x}{k}=\dfrac{\sin 12}{c}$

$\dfrac{\sin x}{k}=\dfrac{\sin 12}{k\sqrt{2-2\cos24}}$

Then apply double angle identity,

$\cos(2\theta)=1-2\sin^2\theta$

$\sin x = \dfrac{\sin 12}{\sqrt{2-2(1-2\sin^2{12})}}$

$=\dfrac{\sin 12}{\sqrt{4\sin^2{12}}}=\dfrac{1}{2}$

$\therefore x=30$

Therefore angle x is 30 degrees.

Remarks:

Some additional beautiful results can be found if we solve x differently, but I will leave this for next time. 🙂

1. Useful info. Hope to see more good posts in the future.

2. Shawn Hicks says:

Notice that this doesn’t work though. You have base angles of 36 and 48 on an ISOSCELES triangle. This doesn’t make sense because base angles are equal. Thus, the given information cannot possibly be true.

• Shawn Hicks says:

NVM.