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2002 AMC 10B Problem 22

2002 AMC 10B Problem 22AMC 10 Step by Step

Name: Date: October 28, 2026

Triangles: Area & Pythagorean · difficulty 4 of 5 · about 4 min · position P21-25

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 $YM = 22$ , find $XY$ .

A) $24$
B) $26$
C) $28$
D) $30$
E) $32$

Answer

Key insight

Call OM = x and ON = y. The two Pythagorean equations are mirror images; adding them gives x^2 + y^2 = 169, and XY is twice 13.

Solution

1. Name the pieces. $M$ is the middle of $OX$ , so call $OM = MX = x$ ; then $OX = 2x$ . Likewise call $ON = NY = y$ ; then $OY = 2y$ .
2. Use the Pythagorean theorem on the two given segments.
- $XN$ is the long side of right triangle $XON$ , whose legs are $OX = 2x$ and $ON = y$ :
$(2x)^2 + y^2 = 19^2, \qquad\text{so}\qquad 4x^2 + y^2 = 361.$
- $YM$ is the long side of right triangle $YOM$ , whose legs are $OM = x$ and $OY = 2y$ :
$x^2 + (2y)^2 = 22^2, \qquad\text{so}\qquad x^2 + 4y^2 = 484.$
3. Decide what we need. $XY$ is the long side of the big triangle with legs $2x$ and $2y$ , so
$XY^2 = 4x^2 + 4y^2 = 4(x^2 + y^2).$
We only need $x^2 + y^2$ , not $x$ and $y$ on their own.
4. Add the two equations. The $4$ s swap places, so adding treats $x$ and $y$ equally:
$5x^2 + 5y^2 = 361 + 484 = 845, \qquad x^2 + y^2 = 169.$
5. So $XY^2 = 4 \cdot 169 = 676$ and $XY = 26$ .

The answer is $\boxed{\textbf{(B)}\ 26}$ .

Why this works

The two equations were mirror images of each other, and the thing we wanted, $x^2 + y^2$ , treats $x$ and $y$ the same way. Adding mirror-image equations is the natural move in that situation. Before solving for each unknown, always check whether the question only needs a combination of them.

Alternative approach

After step 4, notice that $x^2 + y^2$ is the square of $MN$ , since triangle $MON$ has legs $x$ and $y$ . So $MN = 13$ . Triangle $MON$ is the big triangle $XOY$ shrunk by half (both legs are halved), so $XY = 2 \cdot 13 = 26$ .

The trap

Forgetting the legs of the big triangle are 2x and 2y, and answering 13 in your head instead of 26; or grinding out x^2 and y^2 separately and making an arithmetic slip.

Concise solution

Let $OM = MX = x$ and $ON = NY = y$ , so the legs are $OX = 2x$ and $OY = 2y$ .

Segment $XN$ is the hypotenuse of right triangle $XON$ with legs $2x$ and $y$ ; segment $YM$ is the hypotenuse of right triangle $YOM$ with legs $x$ and $2y$ . The Pythagorean theorem gives
$4x^2 + y^2 = 19^2 = 361, \qquad x^2 + 4y^2 = 22^2 = 484.$

The two equations are mirror images, so add them rather than solving:
$5x^2 + 5y^2 = 845 \quad\Longrightarrow\quad x^2 + y^2 = 169.$

The hypotenuse we want satisfies $XY^2 = (2x)^2 + (2y)^2 = 4(x^2 + y^2) = 4 \cdot 169 = 676$ , so $XY = 26$ .

The answer is $\boxed{\textbf{(B)}\ 26}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)

Techniques

Set up the equation/formula and compute; no special trick needed · Exploit symmetry to reduce work or pair up objects