AMC 10 · 2002 · #22
Grade 8 geometry-2dLet △XOY be a right-angled triangle with m∠XOY=90∘. Let M and N be the midpoints of legs OX and OY, respectively. Given that XN=19 and YM=22, find XY.
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: In right triangle $XOY$ the right angle is at $O$, so the two legs are $OX$ and $OY$. Point $M$ is the midpoint of leg $OX$ and point $N$ is the midpoint of leg $OY$. The cross-segment $XN$ (from the far vertex $X$ to the midpoint of the other leg) has length $19$, and $YM$ has length $22$. Find the length of the hypotenuse $XY$.
Givens: $\triangle XOY$ has a right angle at $O$, so $OX \perp OY$.; $M$ is the midpoint of $OX$ and $N$ is the midpoint of $OY$.; $XN = 19$ and $YM = 22$.
Unknowns: The length of the hypotenuse $XY$.
Understand
Restated: In right triangle $XOY$ the right angle is at $O$, so the two legs are $OX$ and $OY$. Point $M$ is the midpoint of leg $OX$ and point $N$ is the midpoint of leg $OY$. The cross-segment $XN$ (from the far vertex $X$ to the midpoint of the other leg) has length $19$, and $YM$ has length $22$. Find the length of the hypotenuse $XY$.
Givens: $\triangle XOY$ has a right angle at $O$, so $OX \perp OY$.; $M$ is the midpoint of $OX$ and $N$ is the midpoint of $OY$.; $XN = 19$ and $YM = 22$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The two given lengths and the answer all depend on the two legs, so the load-bearing move is tool #4 (Introduce a Variable): call the legs $p = OX$ and $q = OY$. Tool #1 (Draw a Diagram) shows that $XN$, $YM$, and $XY$ are each the hypotenuse of a right triangle with its corner at $O$, so the Pythagorean theorem turns every length into an equation in $p$ and $q$. The clever part is tool #7 (Identify Subproblems): notice you never need $p$ and $q$ separately — you only need $p^2 + q^2$, and adding the two equations hands it to you at once.
Execute — Answer: B
6.EE.A.2 Step 1 Name the legs
- Let $p = OX$ and $q = OY$ be the two legs of the right angle at $O$.
- Because $M$ is the midpoint of $OX$, we have $OM = \tfrac{p}{2}$.
- Because $N$ is the midpoint of $OY$, we have $ON = \tfrac{q}{2}$.
- Everything in the problem is now written with just these two letters.
💡 Give the two unknown legs names, and every length in the picture becomes an expression you can compute with.
8.G.B.7 Step 2 Pythagoras on segment XN
- Look at triangle $XON$.
- Its right angle is at $O$ (the same corner), one leg is $OX = p$, and the other leg is $ON = \tfrac{q}{2}$, with $XN$ as the hypotenuse.
- The Pythagorean theorem gives $XN^2 = p^2 + \left(\tfrac{q}{2}\right)^2$.
- Since $XN = 19$, this is $p^2 + \tfrac{q^2}{4} = 361$.
💡 $X$, $O$, and $N$ form a right triangle at $O$, so $XN$ is just its hypotenuse.
8.G.B.7 Step 3 Pythagoras on segment YM
- Now look at triangle $YOM$.
- Its right angle is at $O$ as well, one leg is $OY = q$, and the other leg is $OM = \tfrac{p}{2}$, with $YM$ as the hypotenuse.
- So $YM^2 = q^2 + \left(\tfrac{p}{2}\right)^2$.
- Since $YM = 22$, this is $q^2 + \tfrac{p^2}{4} = 484$.
💡 The same right angle at $O$ makes $YM$ the hypotenuse of a second right triangle.
8.EE.C.8 Step 4 Add the two equations
- Add the two equations.
- On the left, the $p$ terms combine as $p^2 + \tfrac{p^2}{4} = \tfrac{5}{4}p^2$, and the $q$ terms combine the same way as $\tfrac{5}{4}q^2$.
- On the right, $361 + 484 = 845$.
- So $\tfrac{5}{4}\left(p^2 + q^2\right) = 845$.
- Multiply both sides by $\tfrac{4}{5}$ to get $p^2 + q^2 = 676$.
- Notice we found $p^2 + q^2$ without ever solving for $p$ or $q$ alone.
💡 Both equations hide the same bundle $p^2 + q^2$, so adding them pulls that bundle straight out.
8.G.B.7 Step 5 Pythagoras on the hypotenuse XY
- Finally, $XY$ is the hypotenuse of the original right triangle $XOY$, whose legs are $p$ and $q$.
- So $XY^2 = p^2 + q^2 = 676$, which means $XY = \sqrt{676} = 26$.
- The answer is (B).
💡 The very quantity $p^2 + q^2$ that the two equations handed us is exactly $XY^2$.
6.EE.A.2 Let $p = OX$ and $q = OY$ be the two legs of the right angle at $O$. Because $M$ 8.G.B.7 Look at triangle $XON$. Its right angle is at $O$ (the same corner), one leg is 8.G.B.7 Now look at triangle $YOM$. Its right angle is at $O$ as well, one leg is $OY = 8.EE.C.8 Add the two equations. On the left, the $p$ terms combine as $p^2 + \tfrac{p^2}{ 8.G.B.7 Finally, $XY$ is the hypotenuse of the original right triangle $XOY$, whose legs Review
Reasonableness: The answer $26$ is one of the listed choices and sits sensibly among them. As a full check, solve for the legs: subtracting the two equations gives $\tfrac{3}{4}\left(p^2 - q^2\right) = 361 - 484 = -123$, so $p^2 - q^2 = -164$. Combined with $p^2 + q^2 = 676$ this gives $p^2 = 256$ and $q^2 = 420$, i.e. $p = 16$ and $q = \sqrt{420} \approx 20.5$. Check: $XN^2 = 256 + \tfrac{420}{4} = 256 + 105 = 361 = 19^2$ and $YM^2 = 420 + \tfrac{256}{4} = 420 + 64 = 484 = 22^2$ — both match. And $XY = \sqrt{256 + 420} = \sqrt{676} = 26$. Everything is consistent.
Alternative: Use coordinates. Put $O = (0,0)$, $X = (p,0)$, $Y = (0,q)$; then $M = \left(\tfrac{p}{2}, 0\right)$ and $N = \left(0, \tfrac{q}{2}\right)$. The distance formula gives $XN^2 = p^2 + \tfrac{q^2}{4} = 361$ and $YM^2 = q^2 + \tfrac{p^2}{4} = 484$ — the same two equations — and $XY^2 = p^2 + q^2$. Adding still yields $p^2 + q^2 = 676$, so $XY = 26$ (B).
CCSS standards used (min grade 8)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming the legs $p$ and $q$ and writing the half-leg midpoints $\tfrac{p}{2}$ and $\tfrac{q}{2}$.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Turning $XN$, $YM$, and $XY$ into equations, since each is the hypotenuse of a right triangle at $O$.)8.EE.C.8Solve systems of two linear equations by combining them (Adding the two Pythagorean equations to get $p^2 + q^2 = 676$ directly.)
⭐ When two Pythagorean equations both hide the same $p^2 + q^2$, add them instead of solving for each leg — the bundle you want falls right out.
⭐ When two Pythagorean equations both hide the same $p^2 + q^2$, add them instead of solving for each leg — the bundle you want falls right out.
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