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

1961 AHSME Problems/Problem 38

Problem

$\triangle ABC$ is inscribed in a semicircle of radius $r$ so that its base $AB$ coincides with diameter $AB$. Point $C$ does not coincide with either $A$ or $B$. Let $s=AC+BC$. Then, for all permissible positions of $C$:

$\textbf{(A)}\ s^2\le8r^2\qquad \textbf{(B)}\ s^2=8r^2 \qquad \textbf{(C)}\ s^2 \ge 8r^2 \qquad\\ \textbf{(D)}\ s^2\le4r^2 \qquad \textbf{(E)}\ s^2=4r^2$

Solution 1

[asy] draw((-50,0)--(-30,40)--(50,0)--(-50,0)); draw(Arc((0,0),50,0,180)); draw(rightanglemark((-50,0),(-30,40),(50,0),200)); dot((-50,0)); label("A",(-50,0),SW); dot((-30,40)); label("C",(-30,40),NW); dot((50,0)); label("B",(50,0),SE); [/asy] Since $s=AC+BC$, $s^2 = AC^2 + 2 \cdot AC \cdot BC + BC^2$. Since $\triangle ABC$ is inscribed and $AB$ is the diameter, $\triangle ABC$ is a right triangle, and by the Pythagorean Theorem, $AC^2 + BC^2 = AC^2 = (2r)^2$. Thus, $s^2 = 4r^2 + 2 \cdot AC \cdot BC$.


The area of $\triangle ABC$ is $\frac{AC \cdot BC}{2}$, so $2 \cdot [ABC] = AC \cdot BC$. That means $s^2 = 4r^2 + 4 \cdot [ABC]$. The area of $\triangle ABC$ can also be calculated by using base $AB$ and the altitude from $C$. The maximum possible value of the altitude is $r$, so the maximum area of $\triangle ABC$ is $r^2$.


Therefore, $s^2 \le 8r^2$, so the answer is $\boxed{\textbf{(A)}}$.

Solution 2 (Mean Inequality Chain)

First $AB=2r$. $AC^2+BC^2=4r^2$.

By quickly considering extreme cases (when $C$ coincides with $A$ or $B$) we can suspect that the maximum value of $S$ is achieved when $CO$ is perpendicular to $AB$, where $O$ is the origin of the semicircle. We will prove this using the Mean Inequality Chain. Let $AC=x, BC=y$. $x^2+y^2=4r^2$. By QM-AM, $\sqrt{\frac{x^2+y^2}{2}}\ge \frac{x+y}{2}=0.5s$. Plug in $x^2+y^2=4r^2$, we have $2\sqrt{2}r\ge s$. Square both sides, we find that $\boxed{\textbf{(A)}}$ is correct.

~hastapasta

Solution 3 (Calculus Optimization)

This solution utilizes optimization techniques taught in Calculus 1 classes.

Let the equation of the semicircle be $x^2+y^2=r^2, y\ge 0$. Notice that in this semicircle, $y$ is a function of $x$. $y=\sqrt{r^2-x^2}$.

Now let $C(x, \sqrt{r^2-x^2})$. Using the Pythagorean Theorem with $A (-r,0), B(r,0)$, finding $s$ in terms of $x$, differentiating and finding the absolute maximum, we can find that $\boxed{\textbf{(A)}}$ is correct.

P.S.: The process of bash is omitted here since anybody that actually learned optimization problems and paid attention to it (this is core curriculum in regular college math) should be able to bash it out.

~hastapasta

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 37 		Followed by
Problem 39
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
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=1961_AHSME_Problems/Problem_38&oldid=179179"