1971 AHSME Problems/Problem 35

Problem

Each circle in an infinite sequence with decreasing radii is tangent externally to the one following it and to both sides of a given right angle. The ratio of the area of the first circle to the sum of areas of all other circles in the sequence, is

$\textbf{(A) }(4+3\sqrt{2}):4\qquad \textbf{(B) }9\sqrt{2}:2\qquad \textbf{(C) }(16+12\sqrt{2}):1\qquad \\ \textbf{(D) }(2+2\sqrt{2}):1\qquad \textbf{(E) }(3+2\sqrt{2}):1$

Solution 1

$[asy] import geometry; point A = origin; point B = dir(135); point C = (0,sqrt(2)); point D = dir(45); point X = 1/(1+sqrt(2)/2)*(B-C)+C; point Y = 1/(1+sqrt(2)/2)*(D-C)+C; // Circles draw(circle(A,1)); draw(incircle(triangle(C,X,Y))); // Segments XY and BD draw(X--Y); draw(B--D); // Square draw(A--B--C--D--cycle); // Point Labels dot(A); label("A",A,S); dot(B); label("B",B,W); dot(C); label("C",C,N); dot(D); label("D",D,E); dot(X); label("X",X,NW); dot(Y); label("Y",Y,NE); [/asy]$

$\boxed{\textbf{(C) }(16+12\sqrt2):1}$ .

Solution 2 (Informed guess)

Draw a good diagram, and especially make sure that you are drawing a right angle. Then, you will realize that the big circle completely dwarfs all of the other circles. Answer choice (C), $16+12\sqrt2$ , is about $16+12\cdot1.4=32.8$ , whereas the greatest of the other answer choices is about $6.3$ , which is unreasonably small based on our diagram. Thus, we choose answer $\boxed{\textbf{(C) }(16+12\sqrt2):1}$ .

1971 AHSC (Problems • Answer Key • Resources)
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1971 AHSME Problems/Problem 35

Contents

Problem

Solution 1

Solution 2 (Informed guess)

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