2002 AMC 10B Problem 22

Problem 22 of 25HarderAlgebra Geometry

Let

\triangle XOY

be a right-angled triangle with

m\angle XOY = 90^\circ.

Let

M

and

N

be the midpoints of legs

OX

and

OY,

respectively. Given that

XN = 19

and

YM = 22,

what is

XY?

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Solution

Let

OM = a

and

ON = b,

so

OX = 2a

and

OY = 2b.

The right angle at

O

gives

19^2 = (2a)^2 + b^2

and

22^2 = a^2 + (2b)^2.

Adding these,

5(a^2 + b^2) = 19^2 + 22^2 = 845,

so

a^2 + b^2 = 169

and

MN = \sqrt{a^2 + b^2} = 13.

Since

\triangle XOY \sim \triangle MON

with ratio

2,

we have

XY = 2\cdot MN = 26.

Thus, the correct answer is B.

More practice

Concepts: Pythagorean Theorem · median (geometry) · system of equations

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.