AMC 10 · 2002 · #22
Grade 8 geometry-2dPick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 p² + q², and adding the two equations hands it to you at once.
Name the legs
Let p = OX and q = OY be the legs at O. The midpoints give OM = p/2 and ON = q/2, so every length uses just two letters.
Give the two unknown legs names, and every length in the picture becomes an expression you can compute with.
6.EE.A.2Introduce A VariablePythagoras on segment XN
Triangle XON also has its right angle at O, with legs p and q/2 and hypotenuse XN = 19, so p² + q²/4 = 361.
X, O, and N form a right triangle at O, so XN is just its hypotenuse.
8.G.B.7Identify SubproblemsPythagoras on segment YM
Triangle YOM has the same right angle at O, with legs q and p/2 and hypotenuse YM = 22, so q² + p²/4 = 484.
The same right angle at O makes YM the hypotenuse of a second right triangle.
8.G.B.7Identify SubproblemsAdd the two equations
Adding them collects 5/4(p² + q²) = 361 + 484 = 845, and times 4/5 that leaves p² + q² = 676.
Both equations hide the same bundle p² + q², so adding them pulls that bundle straight out.
Both equations hide the same bundle of squared legs, so adding them pulls that bundle straight out.
▸ Why?
Two quantities carrying the identical block can be combined so that only the block survives.
▸ Why?
Each equation comes from a right triangle on the same two legs, which is where that block comes from.
Pythagoras on the hypotenuse XY
XY is the hypotenuse of XOY itself, so XY² = p² + q² = 676 and XY = 26 — choice (B).
The very quantity p² + q² that the two equations handed us is exactly XY².
8.G.B.7Identify SubproblemsWhen two Pythagorean equations both hide the same p² + q², add them instead of solving for each leg — the bundle you want falls right out.
- Name the legs
- Pythagoras on segment XN
- Pythagoras on segment YM
- Add the two equations
- Pythagoras on the hypotenuse XY