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)
(→‎Solution 1)
 
(2 intermediate revisions by the same user not shown)
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 22: Line 22:
 
point X = 1/(1+sqrt(2)/2)*(B-C)+C;
 
point X = 1/(1+sqrt(2)/2)*(B-C)+C;
 
point Y = 1/(1+sqrt(2)/2)*(D-C)+C;
 
point Y = 1/(1+sqrt(2)/2)*(D-C)+C;
 +
 +
circle i = incircle(triangle(C,X,Y));
 +
 +
point R;
 +
 +
// Defining R
 +
pair[] r = intersectionpoints(i,triangle(C,X,Y));
 +
R = r[1];
  
 
// Circles
 
// Circles
 
draw(circle(A,1));
 
draw(circle(A,1));
−
draw(incircle(triangle(C,X,Y)));
+
draw(i);
  
 
// Segments XY and BD
 
// Segments XY and BD
Line 47: Line 55:
 
dot(Y);
 
dot(Y);
 
label("Y",Y,NE);
 
label("Y",Y,NE);
 +
dot(R);
 +
label("R",R,S);
  
 
</asy>
 
</asy>
  
−
<math>\boxed{\textbf{(C) }(16+12\sqrt2):1}</math>.
+
[[WLOG]], let the radius of the largest circle be <math>1</math> (we can do this because scaling the diagram preserves the ratio of areas). Let the largest circle be <math>\omega_1</math> and the second largest circle be <math>\omega_2</math>. As in the diagram above, let <math>\omega_1</math> have center <math>A</math> with points of tangency at <math>B</math> and <math>D</math>. Let the tangents from these two points meet at <math>C</math>. From the problem, we know that <math>\measuredangle BCD=90^{\circ}</math>. Let the common tangent of <math>\omega_1</math> and <math>\omega_2</math> intersect <math>\overline{BC}</math> and <math>\overline{DC}</math> at <math>X</math> and <math>Y</math>, respectively. Let <math>\omega_2</math> intersect <math>\overline{XY}</math> at <math>R</math>.
 +
 
 +
By the [[Two Tangent Theorem]], <math>CB=CD</math>. Thus, <math>\triangle BCD</math> is an isosceles right triangle, so <math>\measuredangle CBD=\measuredangle CDB=45^{\circ}</math>. <math>\measuredangle CBD</math> is a [[tangent-chord angle]] in <math>\omega_1</math>, so <math>\measuredangle BAD=2\cdot\measuredangle CDB=90^{\circ}</math>. Thus, because <math>ABCD</math> has <math>4</math> right angles and <math>2</math> adjacent sides congruent, it is a square. Thus, <math>BC=CD=1</math>.
 +
 
 +
<math>\omega_2</math> is the [[incircle]] of <math>\triangle CXY</math>, because it is tangent to all its sides. Let <math>CX=x</math> and <math>s</math> be the [[semiperimeter]] of <math>\triangle CXY</math>. Because <math>\overline{XY}\parallel\overline{BD}</math>, we see that <math>\measuredangle CXY=\measuredangle CBD=45^{\circ}</math> and <math>\measuredangle CYX=\measuredangle CDB=45^{\circ}</math>. Thus, <math>\triangle CXY</math> is an isosceles right triangle, so <math>XY=x\sqrt2</math>. By the properties of an incircle's touchpoints, <math>XR=s-CY=s-x</math> and <math>YR=s-CX=s-x</math>. Thus, <math>R</math> is the midpoint of <math>\overline{XY}</math>, so <math>RX=XY=\tfrac{\sqrt2}2x</math>.
 +
 
 +
Because <math>\overline{XY}</math> is the common tangent of <math>\omega_1</math> and <math>\omega_2</math>, <math>R</math> is on <math>\omega_1</math>. Thus, by the Two Tangent Theorem, <math>XB=XR=\tfrac{\sqrt2}2x</math>.
 +
 
 +
Now, we have the equation <math>1=BC=BX+XC=\frac{\sqrt2}2x+x</math>, which yields <math>x=\tfrac1{1+\tfrac{\sqrt2}2}=\tfrac{2(2-\sqrt2)}{(2+\sqrt2)(2-\sqrt2)}=2-\sqrt2</math>. We know that <math>s=\tfrac{x+x+x\sqrt2}{2}=x+\tfrac{\sqrt2}2x</math>. Let the [[inradius]] of <math>\triangle CXY</math> be <math>r</math>. Now the formula [[incircle|<math>A=rs</math>]] in <math>\triangle CXY</math> gives us the equation <math>\tfrac{x^2}2=r(x+\tfrac{\sqrt2}2x)</math>, which yields <math>r=\tfrac x{(2+\sqrt2)}=\tfrac{(2-\sqrt2)^2}{(2+\sqrt2)(2-\sqrt2)}=\tfrac{4-4\sqrt2+2}{4-2}=3-2\sqrt2</math>.
 +
 
 +
Because the ratio of the radii of <math>\omega_1</math> and <math>\omega_2</math> is <math>\tfrac{3-2\sqrt{2}}1=3-2\sqrt2</math>, the ratio of their areas is <math>(3-2\sqrt2)^2=9-12\sqrt2+8=17-12\sqrt2</math>. This ratio is the same as that of the area of <math>\omega_2</math> to the third largest circle, the third to the fourth, etc., because removing the largest circle gives a scaled-down version of the same problem. Thus,  the total area of the circles is given by the [[geometric sequence]] starting at <math>\pi</math> (the area of <math>\omega_1</math>) with common ratio <math>17-12\sqrt2\approx17-12\cdot1.4=0.2<1</math>, which, by the infinite [[geometric series]] formula, has an infinite sum of <math>\pi\cdot\tfrac1{1-(17-12\sqrt2)}=\pi\cdot\tfrac{(12\sqrt2+16)}{(12\sqrt2-16)(12\sqrt2+16)}=\pi\cdot\tfrac{3\sqrt2+4}8</math>. However, we are looking for the ratio of the area of <math>\omega_1</math> to that of all of the circles ''excluding'' <math>\omega_1</math>, so we have to subtract <math>\pi</math>. This reveals that the area of the other circles is <math>\pi\cdot\tfrac{3\sqrt2-4}8</math>. Now, our desired answer is the ratio <math>\tfrac{\pi}{\pi\cdot\tfrac{3\sqrt2-4}8}=\tfrac{8(3\sqrt2+4)}{(3\sqrt2-4)(3\sqrt2+4)}=\tfrac{8(3\sqrt2+4)}{18-16}=4(3\sqrt2+4)=12\sqrt2+16</math>. Thus, we choose <math>\boxed{\textbf{(C) }(16+12\sqrt2):1}</math>.
 +
 
 +
== 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), <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 ==

Latest revision as of 20:02, 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; circle i = incircle(triangle(C,X,Y)); point R; // Defining R pair[] r = intersectionpoints(i,triangle(C,X,Y)); R = r[1]; // Circles draw(circle(A,1)); draw(i); // 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); dot(R); label("R",R,S); [/asy]

WLOG, let the radius of the largest circle be $1$ (we can do this because scaling the diagram preserves the ratio of areas). Let the largest circle be $\omega_1$ and the second largest circle be $\omega_2$. As in the diagram above, let $\omega_1$ have center $A$ with points of tangency at $B$ and $D$. Let the tangents from these two points meet at $C$. From the problem, we know that $\measuredangle BCD=90^{\circ}$. Let the common tangent of $\omega_1$ and $\omega_2$ intersect $\overline{BC}$ and $\overline{DC}$ at $X$ and $Y$, respectively. Let $\omega_2$ intersect $\overline{XY}$ at $R$.

By the Two Tangent Theorem, $CB=CD$. Thus, $\triangle BCD$ is an isosceles right triangle, so $\measuredangle CBD=\measuredangle CDB=45^{\circ}$. $\measuredangle CBD$ is a tangent-chord angle in $\omega_1$, so $\measuredangle BAD=2\cdot\measuredangle CDB=90^{\circ}$. Thus, because $ABCD$ has $4$ right angles and $2$ adjacent sides congruent, it is a square. Thus, $BC=CD=1$.

$\omega_2$ is the incircle of $\triangle CXY$, because it is tangent to all its sides. Let $CX=x$ and $s$ be the semiperimeter of $\triangle CXY$. Because $\overline{XY}\parallel\overline{BD}$, we see that $\measuredangle CXY=\measuredangle CBD=45^{\circ}$ and $\measuredangle CYX=\measuredangle CDB=45^{\circ}$. Thus, $\triangle CXY$ is an isosceles right triangle, so $XY=x\sqrt2$. By the properties of an incircle's touchpoints, $XR=s-CY=s-x$ and $YR=s-CX=s-x$. Thus, $R$ is the midpoint of $\overline{XY}$, so $RX=XY=\tfrac{\sqrt2}2x$.

Because $\overline{XY}$ is the common tangent of $\omega_1$ and $\omega_2$, $R$ is on $\omega_1$. Thus, by the Two Tangent Theorem, $XB=XR=\tfrac{\sqrt2}2x$.

Now, we have the equation $1=BC=BX+XC=\frac{\sqrt2}2x+x$, which yields $x=\tfrac1{1+\tfrac{\sqrt2}2}=\tfrac{2(2-\sqrt2)}{(2+\sqrt2)(2-\sqrt2)}=2-\sqrt2$. We know that $s=\tfrac{x+x+x\sqrt2}{2}=x+\tfrac{\sqrt2}2x$. Let the inradius of $\triangle CXY$ be $r$. Now the formula $A=rs$ in $\triangle CXY$ gives us the equation $\tfrac{x^2}2=r(x+\tfrac{\sqrt2}2x)$, which yields $r=\tfrac x{(2+\sqrt2)}=\tfrac{(2-\sqrt2)^2}{(2+\sqrt2)(2-\sqrt2)}=\tfrac{4-4\sqrt2+2}{4-2}=3-2\sqrt2$.

Because the ratio of the radii of $\omega_1$ and $\omega_2$ is $\tfrac{3-2\sqrt{2}}1=3-2\sqrt2$, the ratio of their areas is $(3-2\sqrt2)^2=9-12\sqrt2+8=17-12\sqrt2$. This ratio is the same as that of the area of $\omega_2$ to the third largest circle, the third to the fourth, etc., because removing the largest circle gives a scaled-down version of the same problem. Thus, the total area of the circles is given by the geometric sequence starting at $\pi$ (the area of $\omega_1$) with common ratio $17-12\sqrt2\approx17-12\cdot1.4=0.2<1$, which, by the infinite geometric series formula, has an infinite sum of $\pi\cdot\tfrac1{1-(17-12\sqrt2)}=\pi\cdot\tfrac{(12\sqrt2+16)}{(12\sqrt2-16)(12\sqrt2+16)}=\pi\cdot\tfrac{3\sqrt2+4}8$. However, we are looking for the ratio of the area of $\omega_1$ to that of all of the circles excluding $\omega_1$, so we have to subtract $\pi$. This reveals that the area of the other circles is $\pi\cdot\tfrac{3\sqrt2-4}8$. Now, our desired answer is the ratio $\tfrac{\pi}{\pi\cdot\tfrac{3\sqrt2-4}8}=\tfrac{8(3\sqrt2+4)}{(3\sqrt2-4)(3\sqrt2+4)}=\tfrac{8(3\sqrt2+4)}{18-16}=4(3\sqrt2+4)=12\sqrt2+16$. Thus, we choose $\boxed{\textbf{(C) }(16+12\sqrt2):1}$.

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

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=224781"