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

Difference between revisions of "2005 AIME I Problems/Problem 11"

m (→‎Solution 1)
m (→‎Solution 4)
Line 24: Line 24:
 
=== Solution 4 ===
 
=== Solution 4 ===
 
We proceed by finding the area of the square in 2 different ways. The square is obviously 8*8=64, but we can also find the area in terms of d. From the center of the circle, draw radii that hit the points where the square is tangent to the semicircle. Then the square's area is the area of the small square +2* the area of the trapezoids on the corners+ the area of an isoceles triangle. Adding these all up gives <math>64=\frac{d^2}{4}+\frac{d^2}{4}+(8-\frac{d\sqrt{2}}{2}+\frac{d}{2})(8-\frac{d}{2})</math>
 
We proceed by finding the area of the square in 2 different ways. The square is obviously 8*8=64, but we can also find the area in terms of d. From the center of the circle, draw radii that hit the points where the square is tangent to the semicircle. Then the square's area is the area of the small square +2* the area of the trapezoids on the corners+ the area of an isoceles triangle. Adding these all up gives <math>64=\frac{d^2}{4}+\frac{d^2}{4}+(8-\frac{d\sqrt{2}}{2}+\frac{d}{2})(8-\frac{d}{2})</math>
−
Simplifying gives <math>d-16\sqrt{2}+d\sqrt{2}=0</math>. Solving gives <math>d=32-16\sqrt{2}=32-\sqrt{512}</math> so the answer is 32+512=544
+
Simplifying gives <math>d-16\sqrt{2}+d\sqrt{2}=0</math>. Solving gives <math>d=32-16\sqrt{2}=32-\sqrt{512}</math> so the answer is <math>32+512= \boxed{544}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 18:39, 27 October 2018

Problem

A semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$

Solution

We note that aligning the base of the semicircle with a side of the square is certainly non-optimal. If the semicircle is tangent to only one side of the square, we will have "wiggle-room" to increase its size. Once it is tangent to two adjacent sides of the square, we will maximize its size when it touches both other sides of the square. This can happen only when it is arranged so that the center of the semicircle lies on one diagonal of the square.

2005 AIME I Problem 11.png

Now, let the square be $ABCD$, and let $E \in AB$ and $F \in DA$ be the points at which the "corners" of the semicircle touch the square. Let $O$ be the center of the semicircle.

Solution 1

We can just look at a quarter circle inscribed in a $45-45-90$ right triangle. We can then extend a radius, $r$ to one of the sides creating an $r,r, r\sqrt{2}$ right triangle. This means that we have $r + r\sqrt{2} = 8\sqrt{2}$ so $r = \frac{8\sqrt{2}}{1+\sqrt{2}} = 16 - 8\sqrt{2}$. Then the diameter is $32 - \sqrt{512}$ giving us $32 + 512 = \boxed{544}$

Solution 2

Define the radius of the semicircle as $r$. Draw the perpendicular from $O$ to $AB$, which forms a $45-45-90$ triangle. The length of the perpendicular is $\frac{r}{\sqrt{2}}$. Note also that $AD$ is equal to the length of that perpendicular plus the radius to the point of tangency on $CD$. Thus, $r + \frac{r}{\sqrt{2}} = 8$, and $r = \frac{8\sqrt{2}}{\sqrt{2} + 1} \cdot (\frac{\sqrt{2} - 1}{\sqrt{2} - 1}) = 16 - 8\sqrt{2}$. The diameter is then $2r = 32 - \sqrt{512}$, and the solution is $32 + 512 = 544$.

Solution 3

By the comments above, $AE = AF = a$. By the Pythagorean Theorem, $d^2 = 2a^2$.

Now, if we draw a line through the center, $O$, of the semicircle and its point of tangency with $BC$, we see that this line is perpendicular to $BC$ and so parallel to $AB$. Thus, by triangle similarity it cuts $AF$ in half, and so by symmetry the distance from $O$ to $AD$ is $\frac{a}{2}$ and so the distance from $O$ to $BC$ is $8 - \frac a2$. But this latter quantity is also the radius of the semicircle, so $d = 16 - a$.

Our two previous paragraphs give $2a^2 = (16 - a)^2$ so $a^2 + 32a - 256 = 0$ and $a = 16\sqrt{2} - 16$ (where we discard the negative root of that quadratic) and so $d = a\sqrt{2} = 32 - 16\sqrt{2} = 32 - \sqrt{512}$, so the answer is $32 + 512 = 544$.

Solution 4

We proceed by finding the area of the square in 2 different ways. The square is obviously 8*8=64, but we can also find the area in terms of d. From the center of the circle, draw radii that hit the points where the square is tangent to the semicircle. Then the square's area is the area of the small square +2* the area of the trapezoids on the corners+ the area of an isoceles triangle. Adding these all up gives $64=\frac{d^2}{4}+\frac{d^2}{4}+(8-\frac{d\sqrt{2}}{2}+\frac{d}{2})(8-\frac{d}{2})$ Simplifying gives $d-16\sqrt{2}+d\sqrt{2}=0$. Solving gives $d=32-16\sqrt{2}=32-\sqrt{512}$ so the answer is $32+512= \boxed{544}$.

See also

2005 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=2005_AIME_I_Problems/Problem_11&oldid=98346"