Behold! A plain simple isoseles triangle appears besides you. Frozen in a white boring background this triangle is full of supprises. You must locate the angle by the red lines.

The angle is not 30, it is 20, sorry..

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- Jan 15th 2007, 05:35 PMThePerfectHackerProblem 15
Behold! A plain simple isoseles triangle appears besides you. Frozen in a white boring background this triangle is full of supprises. You must locate the angle by the red lines.

**The angle is not 30, it is 20, sorry.**. - Jan 17th 2007, 10:43 AManthmoo
I only got this far...

- Jan 17th 2007, 10:46 AMCaptainBlack
- Jan 17th 2007, 11:13 AMThePerfectHacker
- Jan 17th 2007, 12:52 PManthmoo
- Jan 17th 2007, 01:03 PMThePerfectHacker
- Jan 18th 2007, 07:50 PMThePerfectHacker
Sorry, the angle should be 20 on top. Not that cannot be solved otherwise, the solution just looks cleaner.

- Jan 18th 2007, 09:24 PMAfterShock
Indeed, that makes it far, far easier. Well, still, this problem took a lot of time and is definitely far harder than it looks.

Any way,

Don't read on if you're still planning on solving this.

Going around the triangle, starting at the bottom left going clockwise, I labelled the triangle A, B, C, where the line going from C intersects AB point D, and the final point I labelled as E. I initially started out by finding all the corresponding angles, just as anthmoo had done. Obviously, they have now changed as a result of the angle change at angle B. I created a perpendicular from AB to point E, and labelled this new point F. Create a bisector from angle A to angle E; parallel transport DE to a point G on BC. Extend a line from angle B to H, and mark H as the intersection of the parallel transported line and the extension of EF. Fill in the corresponding angles by knowing the triangle is an isosceles triangle, and thus the bottom two angles, angles A and C have to add up to 80 each. The other angles are self-evident by supplementary angles and knowing there is 180 degrees in a triangle. We have a kite formed, and we know this as a result of the Angle-Side-Angle postulate comparing triangle BDG and triangle BHG. Connect point D with point G, and thus we have created an isosceles triangle. In fact, we know that triangle ABG is an isosceles triangle, too, since point G is equidistant from both angle A and angle B. Now, looking back at the triangle formed in the kite, we know that those two triangles are congruent because BC is perpendicular to DG, as they are the diagonals of the kite we formed. We now fill in the corresponding angles from the right triangles formed. And finally, we know the other two triangles formed in the kite are congruent by using ASA. After filling in all the corresponding angles and using the fact that there are 180 degrees in a triangle, we now have enough information to determine that angle E must equal 30 degrees.

A diagram would have made this far easier to explain. Nevertheless, very nice problem. - Jan 22nd 2007, 10:43 AMThePerfectHacker
I wish this problem would have been mine, but it is a famous problem from the 1920's I belive. There is an "official" solution involving a construction but I do not know of it. Thus, I will use my solution.

Let $\displaystyle r=AB=BC$.

By the law of cosines,

$\displaystyle AC^2=r^2+r^2-2r^2\cos 20^o=2r^2(1-\cos 20^o)=4r^2\sin^2 10^o$.

Thus,

$\displaystyle AC=2r\sin 10^o$.

Since $\displaystyle AB=BC$, $\displaystyle <A=<C$.

But $\displaystyle <A+<B+<C=180^o$.

$\displaystyle 2<A+20^o=180^o$

$\displaystyle <A=<B=80^o$.

In triangle $\displaystyle AEC$, $\displaystyle <AEC=50^o$.

Thus, $\displaystyle AEC$ is isoseles with $\displaystyle AC=AE$. Thus, $\displaystyle AE=2r\sin 10^o$.

Consider $\displaystyle ADC$, $\displaystyle <ADC=40^o$.

By the law of sines,

$\displaystyle \frac{AC}{\sin 40^o}=\frac{AD}{\sin 80^o}$.

Thus,

$\displaystyle \frac{2r\sin 10^o}{\sin 40^o}=\frac{AD}{2\sin 40^o\cos 40^o}$.

Thus,

$\displaystyle AD=4r\sin 10^o\cos 40^o=4r\sin 10^o\sin 50^o$.

Consider $\displaystyle DEA$, $\displaystyle <EDA=x$ thus, $\displaystyle <DEA=160^o -x$.

By the law of sines,

$\displaystyle \frac{AE}{\sin x}=\frac{AD}{\sin (160^o-x)}$

Thus,

$\displaystyle \frac{2r\sin 10^o}{\sin x}=\frac{4r\sin 10^o\sin 50^o}{\sin (x+20^o)}$

Now we just get to the trigonometry solving.

$\displaystyle 2r\sin 10^o \sin (x+20^o)=4r\sin x \sin 10^o \sin 50^o$ ($\displaystyle r\not = 0$)

$\displaystyle \sin (x+20^o)=2\sin x\sin 50^o$

$\displaystyle \sin x\cos 20^o + \cos x\sin 20^o=2\sin x\sin 50^o$

$\displaystyle \sin x\cos 20^o +\cos x\sin 20^o - 2\sin x\sin 50^o=0$

$\displaystyle \cos x\sin 20^o +\sin x (\cos 20^o-2\sin 50^o)=0$

$\displaystyle \cos x \sin 20^o=\sin x (2\sin 50^o - \cos 20^o)$ ($\displaystyle \cos x \not = 0$).

$\displaystyle \sin 20^o = \tan x (2\sin 50^o - \cos 20^o)$

$\displaystyle \tan x = \frac{\sin 20^o}{2\sin 50^o-\cos 20^o}$

$\displaystyle \tan x = \frac{\sin 20^o}{2\sin (20^o+30^o)-\cos 20^o}$

$\displaystyle \tan x = \frac{\sin 20^o}{2\sin 20^o\cos 30^o+2\sin 30^o\cos 20^0 - \cos 20^0}$

$\displaystyle \tan x = \frac{\sin 20^o}{\sin 20^o \sqrt{3}+\cos 20^o - \cos 20^o}=\frac{\sin 20^o}{\sin 20^o \sqrt{3}}$

$\displaystyle \tan x= \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$

$\displaystyle x=30^o$ - Jan 22nd 2007, 12:01 PMAfterShock
Very nice. I considered using Law of Cosines, but there is far too much trig., and I haven't done trig in years to recall some of the identities.

Good job. - Jan 22nd 2007, 01:31 PMThePerfectHacker
Here is a beautiful site.

Heir.

You can learn a lot from this site.

And it gives the solution without trigonometry.

Though I think my solution is more simpler. :cool: