Difference between revisions of "1959 AHSME Problems/Problem 31"
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== Solution == | == Solution == | ||
| − | <math>\ | + | |
| + | <asy> | ||
| + | |||
| + | import geometry; | ||
| + | |||
| + | point O=(0,0); | ||
| + | real r=5*sqrt(2); | ||
| + | point A=(-r/sqrt(5),2r/sqrt(5)); | ||
| + | point B=(-r/sqrt(5),0); | ||
| + | point C=(5,5); | ||
| + | point D=-C; | ||
| + | |||
| + | markscalefactor=0.1; | ||
| + | |||
| + | dot(O); | ||
| + | label("O",O,S); | ||
| + | |||
| + | // Circle with diameter | ||
| + | draw(circle(O,r)); | ||
| + | draw((-r,0)--(r,0)); | ||
| + | |||
| + | // Small square | ||
| + | draw((-r/sqrt(5),2r/sqrt(5))--(-r/sqrt(5),0)--(r/sqrt(5),0)--(r/sqrt(5),2r/sqrt(5))--(-r/sqrt(5),2r/sqrt(5))); | ||
| + | dot(A); | ||
| + | label("A",A,NW); | ||
| + | dot(B); | ||
| + | label("B",B,SW); | ||
| + | draw(O--A); | ||
| + | draw(rightanglemark(A,B,O)); | ||
| + | |||
| + | // Big Square | ||
| + | draw((5,5)--(5,-5)--(-5,-5)--(-5,5)--(5,5)); | ||
| + | dot(C); | ||
| + | label("C",C,NE); | ||
| + | dot(D); | ||
| + | label("D",D,SW); | ||
| + | draw(C--D); | ||
| + | |||
| + | </asy> | ||
| + | |||
| + | Let the points be labeled as in the diagram, with <math>O</math> being the center of the circle. Because we know that the small square has an area of <math>40</math>, it must have a side length of <math>\sqrt{40}=2\sqrt{10}</math>. It is simple to prove that <math>O</math> is the midpoint of the bottom side of the small square, so <math>OB=\frac{2\sqrt{10}}{2}=\sqrt{10}</math>. By the [[Pythagorean Theorem]], <math>AO=\sqrt{50}=5\sqrt{2}</math>, which is the radius of the circle. Thus, <math>OC=5\sqrt{2}</math>, and so the diagonal <math>\overline{CD}</math> of the big square has length <math>10\sqrt{2}</math>. Thus, the big square has side length <math>\frac{10\sqrt{2}}{\sqrt{2}}=10</math>, and, consequently, it has an area of <math>\boxed{\textbf{(B) }100}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 19:27, 20 July 2024
Problem
A square, with an area of 
, is inscribed in a semicircle. The area of a square that could be inscribed in the entire circle with the same radius, is:
Solution
![[asy] import geometry; point O=(0,0); real r=5*sqrt(2); point A=(-r/sqrt(5),2r/sqrt(5)); point B=(-r/sqrt(5),0); point C=(5,5); point D=-C; markscalefactor=0.1; dot(O); label("O",O,S); // Circle with diameter draw(circle(O,r)); draw((-r,0)--(r,0)); // Small square draw((-r/sqrt(5),2r/sqrt(5))--(-r/sqrt(5),0)--(r/sqrt(5),0)--(r/sqrt(5),2r/sqrt(5))--(-r/sqrt(5),2r/sqrt(5))); dot(A); label("A",A,NW); dot(B); label("B",B,SW); draw(O--A); draw(rightanglemark(A,B,O)); // Big Square draw((5,5)--(5,-5)--(-5,-5)--(-5,5)--(5,5)); dot(C); label("C",C,NE); dot(D); label("D",D,SW); draw(C--D); [/asy]](http://latex.artofproblemsolving.com/b/0/0/b00ac6f64a446e012dc668cb45186d88d34ef702.png)
Let the points be labeled as in the diagram, with 
being the center of the circle. Because we know that the small square has an area of , it must have a side length of 
. It is simple to prove that is the midpoint of the bottom side of the small square, so 
. By the Pythagorean Theorem, 
, which is the radius of the circle. Thus, 
, and so the diagonal 
of the big square has length 
. Thus, the big square has side length 
, and, consequently, it has an area of 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 30 |
Followed by Problem 32 | |
| 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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
