Problem 24 · 2023 AMC 8
Hard
Geometry & Measurement
area-fractionarea-decomposition
Show answer
Answer: A — h = 14.6.
Show hints
Hint 1 of 2
Each cut makes a smaller triangle similar to ABC. The one fact you need: a sub-triangle of height k has area (k/h)2 of ABC — areas scale as the square of height.
Still stuck? Show hint 2 →
Hint 2 of 2
Write both shaded pieces as fractions of ABC and set them equal. Left shaded = whole − top triangle (height 11). Right shaded = the top triangle of height h − 5. The total area T cancels, leaving a clean equation in h.
Show solution
Approach: areas scale as height squared; equate shaded regions
- Both cuts create triangles similar to ABC, so the only tool needed is: area shrinks as the square of height. Crucially, the shaded areas being equal means their fractions of ABC are equal — so the actual area T never matters and cancels out.
- Left figure: the unshaded top triangle has height 11, so it's (11/h)2 of ABC; the shaded trapezoid is the rest, 1 − (11/h)2.
- Right figure: the shaded piece is the top triangle of height h − 5, which is ((h−5)/h)2 of ABC.
- Set equal: 1 − 121/h2 = (h−5)2/h2. Multiply by h2: h2 − 121 = h2 − 10h + 25 — the h2 drops out, giving 10h = 146.
- h = 14.6. Worth keeping: ‘cut parallel to the base’ always makes a similar triangle whose area is (height ratio)2 — convert to fractions of the whole and the unknown total cancels itself away.
Mark:
· log in to save