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

1959 AHSME Problems/Problem 31

Revision as of 19:24, 20 July 2024 by Thepowerful456 (talk | contribs) (→‎Solution)

Problem

A square, with an area of $40$, is inscribed in a semicircle. The area of a square that could be inscribed in the entire circle with the same radius, is: $\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 120\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 200$

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]

Let the points be labeled as in the diagram, with $O$ being the center of the circle. Because we know that the small square has an area of $40$, it must have a side length of $\sqrt{40}=2\sqrt{10}$. It is simple to prove that $O$ is the midpoint of the bottom side of the small square, so $OB=\frac{2\sqrt{10}}{2}=\sqrt{10}$. By the Pythagorean Theorem, $AO=\sqrt{50}=5\sqrt{2}$, which is the radius of the circle. Thus, $OC=5\sqrt{2}$, and so the diagonal $\overline{CD}$ of the big square has length $10\sqrt{2}$. Thus, the big square has side length $\frac{10\sqrt{2}}{\sqrt{2}}=10$, and, subsequently, it has an area of $\boxed{\textbf{(B) }100}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1959_AHSME_Problems/Problem_31&oldid=223233"