Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(diagram)
(informed guess solution)
Line 9: Line 9:
 
\textbf{(E) }(3+2\sqrt{2}):1  </math>
 
\textbf{(E) }(3+2\sqrt{2}):1  </math>
  
−
== Solution ==
+
== Solution 1 ==
  
 
<asy>
 
<asy>
Line 51: Line 51:
  
 
<math>\boxed{\textbf{(C) }(16+12\sqrt2):1}</math>.
 
<math>\boxed{\textbf{(C) }(16+12\sqrt2):1}</math>.
 +
 +
== Solution 2 (Infomred 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), <math>16+12\sqrt2</math>, is about <math>16+12\cdot1.4=32.8</math>, whereas the greatest of the other answer choices is about <math>6.3</math>, which is unreasonably small based on our diagram. Thus, we choose answer <math>\boxed{\textbf{(C) }(16+12\sqrt2):1}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 18:48, 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 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 (Infomred 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 $\boxed{\textbf{(C) }(16+12\sqrt2):1}$.

See Also

1971 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 34 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1971_AHSME_Problems/Problem_35&oldid=224775"