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

2020 AMC 12A Problems/Problem 24

Revision as of 06:38, 2 February 2020 by P groudon (talk | contribs)

Problem 24

Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$, with the property that there is a unique point $P$ inside the triangle such that $AP=1$, $BP=\sqrt{3}$, and $CP=2$. What is $s$?

$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$

Solution

[asy] draw((0,0)--(4,5.65)--(8,0)--cycle); label("A", (4,5.65), N, p = fontsize(10pt)); label("C", (8,0), SE, p = fontsize(10pt)); label("B", (0,0), SW, p = fontsize(10pt)); label("P", (3.5,3.5), NW, p = fontsize(10pt)); draw((0,0)--(3.5,3.5)); label("$\sqrt{3}$",(0,0)--(3.5,3.5), SE); draw((8,0)--(3.5,3.5)); label("2",(8,0)--(3.5,3.5), SW); draw((4,5.65)--(3.5,3.5)); label("1",(4,5.65)--(3.5,3.5), E); draw((8,0)--(4,5.65)--(11,5.65)--cycle); label("$P'$", (6,4.75), NE, p = fontsize(10pt)); draw((4,5.65)--(6,4.75)); label("1",(4,5.65)--(6,4.75), S); draw((8,0)--(6,4.75)); label("$\sqrt{3}$",(8,0)--(6,4.75), E); draw((3.5,3.5)--(6,4.75)); label("1", (3.5,3.5)--(6,4.75), SE); [/asy]

We begin by rotating $\triangle{ABC}$ by $60^{\circ}$ about $A$, such that in $\triangle{A'B'C'}$, $B' = C$. We see that $\triangle{APP'}$ is equilateral with side length $1$, meaning that $\angle APP' = 60^{\circ}$. We also see that $\triangle{CPP'}$ is a $30-60-90$ right triangle, meaning that $\angle CPP'= 60^{\circ}$. Thus, by adding the two together, we see that $\angle APC = 120^{\circ}$. We can now use the law of cosines as following:

\[s^2 = (AP)^2 + (CP)^2 - 2(AP)(CP)\cos{\angle{APC}}\] \[s^2 = 1 + 4 - 2(1)(2)\cos{120^{\circ}}\] \[s^2 = 5 - 4(-\frac{1}{2})\] \[s = \sqrt{5 + 2}\]

giving us that $s = \boxed{\textbf{(B) } \sqrt{7}}$. ~ciceronii

Video Solution

https://www.youtube.com/watch?v=mUW4zcrRL54

See Also

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
All AMC 12 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=2020_AMC_12A_Problems/Problem_24&oldid=116595"