Find the triangle...

Algebra Level 2

If a , b , c ∈ R a,b,c \in \mathbb R are in a geometric progression with common ratio r r , that is b a = c b = r \dfrac ba = \dfrac cb = r , then find b 2 + c 2 a 2 + b 2 \dfrac {b^2+c^2}{a^2+b^2} .

r 2 \frac r2 r r 2 r 2r r 2 r^2 1 r \frac 1r 1 1

Henry U by Henry U , 17, Germany

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2 solutions

Otto Bretscher
Jan 19, 2019

b 2 + c 2 a 2 + b 2 = r 2 a 2 + r 2 b 2 a 2 + b 2 = r 2 \frac{b^2+c^2}{a^2+b^2}=\frac{r^2a^2+r^2b^2}{a^2+b^2}=\boxed{r^2}

2 Helpful 0 Interesting 1 Brilliant 0 Confused
Henry U
Jan 19, 2019

Consider a right triangle with a height of length b b that divides the hypotenuse into two parts of length a a and c c respectively. Let's call its legs d d and e e .

By the geometric mean theorem , b a = c b = r \frac ba = \frac cb = r and by similarity of triangles and the pythagorean theorem, r = b a = e d = b 2 + c 2 a 2 + b 2 ⇔ b 2 + c 2 a 2 + b 2 = r 2 r = \frac ba = \frac ed = \frac {\sqrt{b^2+c^2}}{\sqrt{a^2+b^2}} \Leftrightarrow \frac {b^2+c^2}{a^2+b^2} = \boxed{r^2} .

1 Helpful 1 Interesting 0 Brilliant 0 Confused

Interesting approach!!!

Aaghaz Mahajan - 2 years, 4 months ago

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