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Difference between revisions of "1972 AHSME Problems/Problem 23"

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\textbf{(C) }1.25\qquad
 
\textbf{(C) }1.25\qquad
 
\textbf{(D) }\frac{5\sqrt{17}}{16}\qquad  
 
\textbf{(D) }\frac{5\sqrt{17}}{16}\qquad  
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\textbf{(E) }\text{None of these}</math>  
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\textbf{(E) }\text{None of these}</math>
  
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//draw((0,0)--(0,1)--(2,1)--(2,0)--cycle^^(.5,1)--(.5,2)--(1.5,2)--(1.5,1)--cycle^^(1,2)--(1,0)--cycle^^(.5,2)--(1,13/16)--);
 
 
==Solution 1==
 
==Solution 1==
 
<asy>
 
<asy>

Revision as of 00:14, 26 May 2024

Problem 23

[asy] draw((0,0)--(0,1)--(2,1)--(2,0)--cycle^^(.5,1)--(.5,2)--(1.5,2)--(1.5,1)--cycle^^(1,2)--(1,0)); [/asy]

The radius of the smallest circle containing the symmetric figure composed of the 3 unit squares shown above is

$\textbf{(A) }\sqrt{2}\qquad \textbf{(B) }\sqrt{1.25}\qquad \textbf{(C) }1.25\qquad \textbf{(D) }\frac{5\sqrt{17}}{16}\qquad \textbf{(E) }\text{None of these}$

Solution 1

[asy] size(180); defaultpen(fontsize(10pt)); pair A =origin; pair B = (0,1); pair C = (2,1); pair D = (2,0); pair E = (1,0); pair F = (1,2); pair G = (0.5,2); pair H = (1.5,2); pair I = (1.5,1); pair J = (0.5,1); pair K = (1,13/16); label("$A$",A,SW); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,SE); label("$E$",E,S); label("$F$",F,SE); label("$G$",G,NW); label("$H$",H,NE); label("$I$",I,NW); label("$J$",J,SW); label("$K$",K,SE); label("$1$",(0.5,0),S); label("$1$",(0,0.5),W); label("$1$",(0.5,1.5),W); label("$\frac{1}{2}$",(0.75,2),S); label("$x$",(1,13/32),E); label("$2-x$",(1,45/32),E); label("$r$",((A+K)/2),SE); label("$r$",((G+K)/2),SW); draw(A--B--C--D--E--F--G--H--I--J--G--F--E--A--K--G); draw(arc(K,5*sqrt(17)/16,0,360)); [/asy]

Draw lines from points $A$ and $G$ to the center of the circle $K$. As shown from the diagram, $r$ is the radius of the circumscribed circle. Set $\overline{EK}$ to $x$, so as $\overline{EF}$ is $1+1=2$, $\overline{FK}$ is $2-x$. Then, $\triangle FGK$ and $\triangle AEK$ are both right, so an equation can be formed with the Pythagorean Theorem. The radii (hypotenuses) are equal, so the following equation can be made and solved:\\ \begin{align*} \frac{1}{2}^2+(2-x)^2&=1^2+x^2\\ \frac{1}{4}+(-x+2)^2&=1+x^2\\ \frac{1}{4}+x^2-4x+4&=1+x^2\\ \frac{1}{4}-4x+4&=1\\ -4x&=-3-\frac{1}{4}\\ 4x&=\frac{13}{4}\\ x&=\frac{13}{16}\\ \end{align*} $x$ is solved, so plugging it in to the original formula yields: \begin{align*} \sqrt{1^2+\frac{13}{16}^2} \\ =\sqrt{1+\frac{169}{256}} \\ =\sqrt{\frac{425}{256}} \\ =\frac{\sqrt{425}}{\sqrt{256}} \\ =\frac{5\sqrt{17}}{16} \\ =\boxed{\textbf{(D)}}. \end{align*} ~airbus-a321

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