AMC 10 · 2002 · #20
Grade 8 geometry-2dPick an answer.
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
Name the legs p and q; the midpoints then sit at p/2 and q/2.
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
Pythagoras on the first cross-segment gives 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
The second gives q² + p²/4 = 484, the mirror image.
The same right angle at O makes YM the hypotenuse of a second right triangle.
8.G.B.7Identify SubproblemsAdd the two equations
Adding combines them into five quarters of the total, giving 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 two squares, so adding them pulls that bundle straight out.
▸ Why?
The right angle at the centre makes each segment the hypotenuse of a right triangle, so its square is the two legs squared added.
▸ Why?
Adding two true equations keeps them true, so the shared bundle can be collected without solving for either leg.
Pythagoras on the hypotenuse XY
That sum is the hypotenuse squared, so 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