2002 AMC 10B Problem 22

Below is the professionally curated solution for Problem 22 of the 2002 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10B solutions, or check the answer key.

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Concepts:Pythagorean Theoremmedian (geometry)system of equations

Difficulty rating: 1690

22.

Let XOY\triangle XOY be a right-angled triangle with mXOY=90.m\angle XOY = 90^\circ. Let MM and NN be the midpoints of legs OXOX and OY,OY, respectively. Given that XN=19XN = 19 and YM=22,YM = 22, what is XY?XY?

2424

2626

2828

3030

3232

Solution:

Let OM=aOM = a and ON=b,ON = b, so OX=2aOX = 2a and OY=2b.OY = 2b. The right angle at OO gives 192=(2a)2+b2and222=a2+(2b)2.19^2 = (2a)^2 + b^2 \quad\text{and}\quad 22^2 = a^2 + (2b)^2.

Adding these, 5(a2+b2)=192+222=845,5(a^2 + b^2) = 19^2 + 22^2 = 845, so a2+b2=169a^2 + b^2 = 169 and MN=a2+b2=13.MN = \sqrt{a^2 + b^2} = 13.

Since XOYMON\triangle XOY \sim \triangle MON with ratio 2,2, we have XY=2MN=26.XY = 2\cdot MN = 26.

Thus, the correct answer is B.

Problem 22 in Other Years