Problem 23 · 1999 AMC 8
Stretch
Geometry & Measurement
area-to-lengthpythagorean
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Answer: C — √13.
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Hint 1 of 2
You want a length, but you're given an area — so use area to FIND the length first. Triangle BMC is right-angled at B with one leg the full side BC = 3; its area being one-third of the square pins down the other leg BM.
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Hint 2 of 2
Once you know both legs of the right triangle BMC, segment CM is just its hypotenuse — Pythagoras.
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Approach: let the equal-area condition solve for the missing leg, then Pythagoras
- The square's area is 3² = 9, split into three equal parts of 3 each. Triangle BMC is right-angled at corner B, with legs BC = 3 (top side) and BM along the left side. Its area = ½ · 3 · BM = 3, so BM = 2.
- Now CM is the hypotenuse of right triangle BMC: CM = √(BM² + BC²) = √(2² + 3²) = √(4 + 9) = √13.
- Why this transfers: when a problem hands you an area but asks a length, run the area formula backward to recover the unknown side — then geometry (here Pythagoras) finishes it. The √13 answer also tells you it's between √9 = 3 and √16 = 4, a quick sanity check on the choices.
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