Difference between revisions of "1971 AHSME Problems/Problem 35"

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== Solution ==
 
== Solution ==
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<asy>
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import geometry;
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point A = origin;
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point B = dir(135);
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point C = (0,sqrt(2));
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point D = dir(45);
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point X = 1/(1+sqrt(2)/2)*(B-C)+C;
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point Y = 1/(1+sqrt(2)/2)*(D-C)+C;
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// Circles
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draw(circle(A,1));
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draw(incircle(triangle(C,X,Y)));
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// Segments XY and BD
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draw(X--Y);
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draw(B--D);
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// Square
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draw(A--B--C--D--cycle);
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// Point Labels
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dot(A);
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label("A",A,S);
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dot(B);
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label("B",B,W);
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dot(C);
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label("C",C,N);
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dot(D);
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label("D",D,E);
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dot(X);
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label("X",X,NW);
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dot(Y);
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label("Y",Y,NE);
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</asy>
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<math>\boxed{\textbf{(C) }(16+12\sqrt2):1}</math>.
 
<math>\boxed{\textbf{(C) }(16+12\sqrt2):1}</math>.
  

Revision as of 18:38, 8 August 2024

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

[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]

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

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 34
Followed by
Last Problem
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