They Look Like Pythagorean Triple

Geometry Level 4

Triangle A B C ABC above is an equilateral triangle and A D = 3 AD=3 , B D = 4 BD=4 and C D = 5 CD=5 . Find A B AB .


The answer is 6.76643.

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5 solutions

Ahmad Saad
May 16, 2016

Nice solution.

Niranjan Khanderia - 5 years, 1 month ago

Amazing solution.

Adarsh Adi - 1 year, 11 months ago

link text
In an equivalent triangle side S, if a point inside it is at distances X, Y, Z from the vertices, their relation is given in the link above by,
3 ( X 4 + Y 4 + Z 4 + S 4 ) = ( X 2 + Y 2 + Z 2 + S 2 ) 2 3 ( 3 4 + 4 4 + 5 4 + S 4 ) = ( 3 2 + 4 2 + 5 2 + S 2 ) 2 . E x p a n d i n g a n d s i m p l i f y i n g , 2 S 2 100 S + 386 = 0 o r S 2 50 S + 193 = 0. S o l v i n g t h e q u a d r a t i c S 2 = 45.7846 , r e j e c t i n g t h e o t h e r r o o t b e i n g t o o s m a l l . S = 6.7664325 3(X^4+Y^4+Z^4+S^4)=(X^2+Y^2+Z^2+S^2)^2 \\ \implies\ 3(3^4+4^4+5^4+S^4)=(3^2+4^2+5^2+S^2)^2.\\ Expanding\ and\ simplifying,\\ 2S^2-100S+386=0\ \ or\ \ S^2 - 50S+193=0.\\ Solving\ the\ quadratic\ \ S^2=45.7846,\ rejecting \ the\ other\ root\ being\ too\ small.\\ \therefore\ S=\color{#D61F06}{6.7664325}

Nice, I did not know the result. Thanks for sharing your solution

Sathvik Acharya - 4 years, 2 months ago

Refer here to get that twice the area of the equilateral triangle is 25 3 2 + 18 25\frac{\sqrt3}{2}+18 Hence, if the side of the required triangle is a a , a 2 3 2 = 25 3 2 + 18 a = 25 + 12 3 6.766 a^2 \frac{\sqrt3}{2} = 25\frac{\sqrt3}{2}+18\\\implies a=\sqrt{25+12\sqrt3} \approx 6.766

Rotate triangle A B C ABC in magnitude of 6 0 60^\circ in the counterclockwise direction as shown in my figure.

sin D D C = 3 5 \sin \angle D’DC=\dfrac{3}{5}

D D C 36.8 7 \angle D’DC \approx 36.87^\circ

By cosine rule, we have

( B C ) 2 = 4 2 + 5 2 2 ( 4 ) ( 5 ) [ cos ( 60 + 36.87 ) ] (BC)^2=4^2+5^2-2(4)(5)[\cos (60+36.87)]

B C = A B 6.766 units BC = AB \approx 6.766~\text{units}

Paola Ramírez
May 21, 2016

Let generalize this problem for any Pythagorean triple with cathetus a a and b b , and hypotenuse c c .

First moves A P C \triangle APC together A P B \triangle APB as image shows, clearly, A P P \triangle AP'P is an equilateral triangle and P P B \triangle PP'B is an rectangle triangle of sides a , b a,b and c c so B P A = 60 ° + 90 ° = 150 ° \angle BPA = 60° + 90° = 150° .

By law of cosines, the side of the equilateral triangle is s = a 2 + b 2 2 ( a ) ( b ) cos 150 ° \boxed{s=\sqrt{a^2 + b^2 - 2(a)(b)\cos 150°}} .

For this triangle A B = 3 2 + 4 2 2 ( 3 ) ( 4 ) cos 150 ° 6.766 \boxed{AB=\sqrt{3^2+4^2-2(3)(4)\cos150°}\approx6.766} .

Similar problem: Equilateral triangle point

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