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 | ||
| − | \textbf{(E) }\text{None of these}</math> | + | \textbf{(E) }\text{None of these}</math> |
| − | |||
==Solution 1== | ==Solution 1== | ||
<asy> | <asy> | ||
| Line 52: | Line 51: | ||
</asy> | </asy> | ||
| − | Draw lines from points <math>A</math> and <math>G</math> to the center of the circle <math>K</math>. As shown from the diagram, <math>r</math> is the radius of the circumscribed circle. Set <math>\overline{EK}</math> to <math>x</math>, so as <math>\overline{EF}</math> is <math>1+1=2</math>, <math>\overline{FK}</math> is <math>2-x</math>. Then, <math>\triangle FGK</math> and <math>\triangle AEK</math> 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: | + | Draw lines from points <math>A</math> and <math>G</math> to the center of the circle <math>K</math>. As shown from the diagram, <math>r</math> is the radius of the circumscribed circle. Set <math>\overline{EK}</math> to <math>x</math>, so as <math>\overline{EF}</math> is <math>1+1=2</math>, <math>\overline{FK}</math> is <math>2-x</math>. Then, <math>\triangle FGK</math> and <math>\triangle AEK</math> 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: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\frac{1}{2}^2+(2-x)^2&=1^2+x^2\\ | \frac{1}{2}^2+(2-x)^2&=1^2+x^2\\ | ||
Latest 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]](http://latex.artofproblemsolving.com/0/4/7/047621775f4f75ae22b66f7678a70ef091bc4943.png)
The radius of the smallest circle containing the symmetric figure composed of the 3 unit squares shown above is
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]](http://latex.artofproblemsolving.com/e/9/3/e938fa0f0a6452d862f2a73396475c4b35bd60c2.png)
Draw lines from points 
and 
to the center of the circle 
. As shown from the diagram, 
is the radius of the circumscribed circle. Set 
to 
, so as 
is 
, 
is 
. Then, 
and 
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:
is solved, so plugging it in to the original formula yields:
~airbus-a321
