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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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  • $\begingroup$ What is the source? $\endgroup$ – Sawarnik Feb 22 '14 at 12:33
  • $\begingroup$ @Sawarnik By Trigonometry can be programmed. $\endgroup$ – felipeuni Feb 25 '14 at 0:45
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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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  • $\begingroup$ If some one could remove the <!-- ... > in the last line so that the image is displayed, I'd appreciate that. Thanks! $\endgroup$ – none Oct 30 '12 at 13:08
  • $\begingroup$ Agree. It can be proved (beaten) by trigonometry for sure, but using a pure geometric approach one can also reveal the hidden relations in the picture and so explain that beautiful formula. $\endgroup$ – Jeyekomon Oct 30 '12 at 13:33
  • $\begingroup$ @none, can you tell me what software did you use to make that diagram? $\endgroup$ – Swair Oct 31 '12 at 5:59
  • $\begingroup$ @swair, I'm pretty sure it's GeoGebra. $\endgroup$ – Rahul Nov 4 '12 at 5:39
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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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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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