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1972 AMC 12 Problem 32

Attempt Problem 32 of the 1972 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1972 AMC 12 solutions, or check the answer key.

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32.

Chords ABAB and CDCD in the circle shown intersect at EE and are perpendicular to each other. If segments AE,AE, EB,EB, and EDED have measures 2,2, 6,6, and 33 respectively, then the length of the diameter of the circle is:

454\sqrt5

65\sqrt{65}

2172\sqrt{17}

373\sqrt7

626\sqrt2

Answer: B
Concepts:power of a pointcoordinate geometrydistance formulacircle
Difficulty rating: 1890
Small Hint:

First use AE⋅EB=CE⋅EDAE\cdot EB=CE\cdot ED

Big Hint:

Place EE at the origin; the center is at the intersection of the two chord perpendicular bisectors

Solution:

The intersecting-chords theorem gives 2⋅6=CE⋅3, 2\cdot6=CE\cdot3, so CE=4.CE=4. Put E=(0,0),E=(0,0), A=(−2,0),A=(-2,0), B=(6,0),B=(6,0), C=(0,4),C=(0,4), and D=(0,−3).D=(0,-3). The perpendicular bisectors of ABAB and CDCD meet at O=(2,12). O=\left(2,\frac12\right). Thus r2=OA2=42+(12)2=654, r^2=OA^2=4^2+\left(\frac12\right)^2=\frac{65}{4}, and the diameter is 2r=65.2r=\sqrt{65}.

Therefore, the correct answer is B.

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