Ratio Geometry Problem

Geometry Level 3

A square is drawn in the corner of a right-triangle as shown. Let a = B Q , b = B P , c = P Q a = BQ, b = BP, c = PQ .

Which expression gives the ratio of the area of the triangle (minus the area of the square) to the area of the square ?

1 : 1 1:1 a b : c 2 ab : c^{2} c 2 : 2 a b c^{2} : 2ab c : a + b c : a+b

Connor McCartney by Connor McCartney , 18, Australia

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1 solution

Connor McCartney
Jan 7, 2018

let x = the side-length of the square

Triangles PFE and PBQ are similar (AAA).

Thus by similarity,

b − x x \frac{b-x}{x} = b a \frac{b}{a}

a (b - x) = bx

ab - ax = bx

x (a + b) = ab

x = a b a + b \frac{ab}{a+b}

Now we solve for the ratio:

a r e a o f t r i a n g l e m i n u s a r e a o f s q u a r e a r e a o f s q u a r e \frac{area of triangle minus area of square}{area of square} = ( 1 / 2 ) b h − s 2 s 2 \frac{(1/2)bh - s^2}{s^2}

= ( a b / 2 ) − x 2 x 2 \frac{(ab/2)-x^2}{x^2}

= a b − 2 x 2 2 \frac{ab-2x^2}{2} / x^2

= a b − 2 x 2 2 x 2 \frac{ab-2x^2}{2x^2}

now substitute x = 2 a b a + b \frac{2ab}{a+b}

= a b − 2 ( 2 a b / a + b ) 2 2 ( 2 a b / a + b ) 2 \frac{ab-2(2ab/a+b)^2}{2(2ab/a+b)^2}

= (ab - 2 ( a 2 ) ( b 2 ) ( a + b ) 2 \frac{2(a^2)(b^2)}{(a+b)^2} ) / 2 ( a 2 ) ( b 2 ) ( a + b ) 2 \frac{2(a^2)(b^2)}{(a+b)^2}

= a b ( a + b ) 2 = a b ( 2 a b ) a b ( 2 a b ) \frac{ab(a+b)^2 = ab(2ab)}{ab(2ab)}

= ( a + b ) 2 − 2 a b 2 a \frac{(a+b)^2 - 2ab}{2a}

= a 2 + b 2 2 a b \frac{a^2 + b^2}{2ab}

substitute c^2 = a^2 + b^2 (pythagoras)

= c^2 : 2ab

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