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Geometry: Buildings in the triangle enter image description here Other triangles with the same property:

$1.$ 12 18 6 12 30 102

$2.$ 15 30 15 15 15 90

$3.$ 24 30 54 24 6 42

$4.$ 30 10 40 30 20 50 (proposed problem in sense of clockwise)

$5.$ 36 12 6 12 18 96

$6.$ 36 18 6 36 6 78

$7.$ 42 6 36 42 12 42

$8.$ 60 6 57 30 3 24

$9.$ 60 24 12 12 6 66

Using matlab we can find all triangles (integer solutions) with this property sums of squares:

enter image description here

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    @Sawarnik By Trigonometry can be programmed.2014-02-25

3 Answers 3

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Denote the length of the lower side (the hypotenuse of the big triangle) by $h$. Then the side containing $c$ is of length $h\sin 40^\circ=h\cos 50^\circ$ (Do you see why?)
Now let's find the length of segment from the rightmost bottom vertex to the one at the bottom of $b$ (denote it by $l_1$): Using the law of sines, we get $\frac{h\sin 40^\circ}{\sin 110^\circ}=\frac{l_1}{\sin20^\circ}$, so $l_1=\frac{h\sin 40^\circ\sin20^\circ}{\sin 110^\circ}$
Denote by $l_2$ the length of the segment between the bottom of $b$ and the bottom of $a$. Then $\frac{l_1+l_2}{\sin50^\circ}=\frac{h\sin 40^\circ}{\sin 80^\circ}$, so $l_2=\frac{h\sin 40^\circ\sin50^\circ}{\sin 80^\circ}-l_1=h\sin 40^\circ\left(\frac{\sin50^\circ}{\sin 80^\circ}-\frac{\sin20^\circ}{\sin 110^\circ}\right)$ Now, we can express $a,b,c$ in terms of $h$: $\frac{a}{\sin30^\circ}=\frac{h-l_1-l_2}{\sin(180^\circ-30^\circ-(180^\circ-40^\circ-40^\circ))}=\frac{h-l_1-l_2}{\sin50^\circ} \\ \frac{b}{\sin30^\circ}=\frac{h-l_1}{\sin(180^\circ-30^\circ-(180^\circ-40^\circ-70^\circ))}=\frac{h-l_1}{\sin80^\circ} \\ \frac{c}{\sin30^\circ}=\frac{h}{\sin(180^\circ-50^\circ-30^\circ)}=\frac{h}{\sin100^\circ}$ Now you can simplify all those expressions and substitute. (To calculate the angles, I used the fact that the sum of angles in a triangle is $180^\circ$ and each time I looked at a different triangle)

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I add the letters to your points

Original graph with letters

Using Theorem of Sine, we get $\frac{a}{\sin 30^\circ}=\frac{BD}{\sin(10^\circ+40^\circ)}=\frac{BD}{\sin 50^\circ}$ $\frac{BD}{\sin 40^\circ}=\frac{BA}{\sin(30^\circ+20^\circ+50^\circ)}=\frac{BA}{\sin 100^\circ}$ so we get $a=\left(\frac{BA\cdot\sin 30^\circ}{\sin 100^\circ\sin 50^\circ}\right)\cdot\sin 40^\circ$ Also we have $\frac{b}{\sin 30^\circ}=\frac{BE}{\sin(10^\circ+40^\circ+30^\circ)}=\frac{BE}{\sin 80^\circ}$ because $\angle BAE=\angle BEA=70^\circ$, we have $BE=BA$ so we get $b=\frac{BA\cdot\sin 30^\circ}{\sin 80^\circ}=\frac{BA\cdot\sin 30^\circ\cdot\sin 50^\circ}{\sin 100^\circ\cdot\sin 50^\circ}=\left(\frac{BA\cdot\sin 30^\circ}{\sin 100^\circ\sin 50^\circ}\right)\cdot\cos 40^\circ$ Finally, we have $\frac{c}{\sin 30^\circ}=\frac{BC}{\sin(10^\circ+40^\circ+30^\circ+20^\circ)}=\frac{BC}{\sin 100^\circ}$ $BC=\frac{BA}{\sin 50^\circ}$ So, we have $c=\frac{BA\cdot\sin 30^\circ}{\sin 100^\circ\sin 50^\circ}$

Since $\sin^2 40^\circ+\cos^2 40^\circ=1$

So we have $a^2+b^2=c^2$

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I thought that this problem should be solvable without using trigonometry. Here's a hint for a geometric solution:

enter image description here

Draw the segment $DG$ and let $B'$ be the intersection of the line through $AE$ and the perpendicular line to $DG$. I claim that the triangle $DB'G$ has sides of length $a,b,c$, so $c^2 = a^2 + b^2$ by the Pythagorean theorem.

Since geometry is hard to communicate, I think it is better to let you figure this out on your own.

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    @swair, I'm pretty sure it's [GeoGebra](http://www.geogebra.org/cms/).2012-11-04