High School Challenge 3

Algebra Level 3

In the equation ( x − y ) ( x − z ) (x-y)(x-z) , y y and z z are the shorter sides of a right-angled triangle where x x is a real number and the hypotenuse is of length b 2 + 2 b c \sqrt{b^{2}+2bc} . If ( x − y ) ( x − z ) = x 2 + b x + c (x-y)(x-z) = x^{2}+bx+c (where b and c are real numbers), find y ( y − 1 ) z ( z − 1 ) \frac{y(y-1)}{z(z-1)} .


The answer is 1.

Steven Lee by Steven Lee , 22, Indonesia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Steven Lee
Sep 17, 2014

y 2 + z 2 = b 2 + 2 b c y^{2}+z^{2} = b^{2}+2bc

x 2 + b x + c = x 2 − x y − x z + y z = x 2 + ( − ) ( y + z ) x + y z x^{2}+bx+c = x^{2}-xy-xz+yz = x^{2}+(-)(y+z)x+yz

b = − ( y + z ) , c = y z b = -(y+z), c = yz

Thus

y 2 + z 2 = ( − ( y + z ) ) 2 + 2 y z ( − ( y + z ) ) = y 2 + 2 y z + z 2 + 2 y z ( − ( y + z ) ) y^{2}+z^{2} = (-(y+z))^{2} + 2yz(-(y+z)) = y^{2}+2yz+z^{2}+2yz(-(y+z))

2 y z + 2 y z ( − ( y + z ) ) = 0 2yz+2yz(-(y+z)) = 0

2 y z = − 2 y z ( − ( y + z ) ) 2yz = -2yz(-(y+z))

1 = y + z 1 = y+z

y 2 − z 2 = ( y + z ) ( y − z ) = y − z y^{2}-z^{2} = (y+z)(y-z) = y-z

y 2 − y = z 2 − z y^{2}-y = z^{2}-z

y ( y − 1 ) z ( z − 1 ) = 1 \frac{y(y-1)}{z(z-1)} = 1

3 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...