2024 AIME I Problem 5

Below is the professionally curated solution for Problem 5 of the 2024 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AIME I solutions, or check the answer key.

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Concepts:coordinate geometrycircleperpendicular bisector

Difficulty rating: 2390

5.

Rectangle ABCDABCD has dimensions AB=107AB = 107 and BC=16,BC = 16, and rectangle EFGHEFGH has dimensions EF=184EF = 184 and FG=17.FG = 17. Points D,D, E,E, C,C, and FF lie on line DFDF in that order, and AA and HH lie on opposite sides of line DF,DF, as shown. Points A,A, D,D, H,H, and GG lie on a common circle. Find CE.CE.

Solution:

Put line DFDF on the xx-axis with D=(0,0)D = (0, 0) and C=(107,0),C = (107, 0), so A=(0,16).A = (0, -16). Let DE=e.DE = e. Then E=(e,0),E = (e, 0), F=(e+184,0),F = (e + 184, 0), and the second rectangle sits above the line: H=(e,17)H = (e, 17) and G=(e+184,17).G = (e + 184, 17).

The center of the circle through A,A, D,D, H,H, GG lies on the perpendicular bisector of the vertical segment AD,\overline{AD}, the line y=8,y = -8, and on the perpendicular bisector of the horizontal segment HG,\overline{HG}, the line x=e+92.x = e + 92. Equating the center's squared distances to DD and to H,H, (e+92)2+82=922+252=9089,(e + 92)^2 + 8^2 = 92^2 + 25^2 = 9089, so (e+92)2=9025(e + 92)^2 = 9025 and e+92=95,e + 92 = 95, giving e=3.e = 3.

Therefore CE=DCDE=1073=104.CE = DC - DE = 107 - 3 = 104.

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