1952 AMC 12 Problem 29

Attempt Problem 29 of the 1952 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 1952 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

29.

In a circle of radius 55 units, CDCD and ABAB are perpendicular diameters. A chord CHCH cutting ABAB at KK is 88 units long. The diameter ABAB is divided into two segments whose dimensions are:
1.25,1.25, 8.758.75
2.75,2.75, 7.257.25
2,2, 88
4,4, 66
None of these
Answer: A
Concepts:circlechordcoordinate geometry
Small hint:
Put the center at the origin, take C=(0,5)C=(0,5), and represent the other endpoint HH of the chord by coordinates
Big hint:
Use CH=8CH=8 and OH=5OH=5, then find where line CHCH crosses the horizontal diameter
Solution:
Put the circle at the origin with C=(0,5)C=(0,5) and let H=(x,y).H=(x,y). The equations x2+y2=25,x2+(y−5)2=64 \begin{aligned} x^2+y^2&=25,\\ x^2+(y-5)^2&=64 \end{aligned} give y=−75y=-\frac{7}{5} and x=±245.x=\pm\frac{24}{5}. Taking x=245x=\frac{24}{5} does not change the segment lengths.
The line from C=(0,5)C=(0,5) to H=(245,−75)H=(\frac{24}{5},-\frac{7}{5}) crosses y=0y=0 at x=154.x=\frac{15}{4}. Its distances to the endpoints (−5,0)(-5,0) and (5,0)(5,0) of diameter ABAB are therefore 5+154=354=8.75,5−154=54=1.25. \begin{aligned} 5+\frac{15}{4}&=\frac{35}{4}=8.75,\\ 5-\frac{15}{4}&=\frac54=1.25. \end{aligned}
Thus, the correct answer is A.
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