1971 AHSME Problems/Problem 35

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 34
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