Difference between revisions of "2020 AMC 12A Problems/Problem 24"

Revision as of 19:31, 1 February 2020

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

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Difference between revisions of "2020 AMC 12A Problems/Problem 24"

Revision as of 19:31, 1 February 2020

Problem 24

Solution

@@ Line 4: / Line 4: @@
 <math>\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}</math>
-==Solution 1==
+==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 <math>\triangle{ABC}</math> by <math>60^{\circ}</math> about <math>A</math>, such that in <math>\triangle{A'B'C'}</math>, <math>B' = C</math>. We see that
+<math>\triangle{APP'}</math> is equilateral with side length <math>1</math>, meaning that <math>\angle APP' = 60^{\circ}</math>. We also see that <math>\triangle{CPP'}</math> is a <math>30-60-90</math> right triangle, meaning that <math>\angle CPP'= 60^{\circ}</math>. Thus, by adding the two together, we see that <math>\angle APC = 120^{\circ}</math>. We can now use the law of cosines as following:
+<cmath>s^2 = (AP)^2 + (CP)^2 - 2(AP)(CP)\cos{\angle{APC}}</cmath>
+<cmath>s^2 = 1 + 4 - 2(1)(2)\cos{120^{\circ}}</cmath>
+<cmath>s^2 = 5 - 4(-\frac{1}{2})</cmath>
+<cmath>s = \sqrt{5 + 2}</cmath>
+giving us that <math>s = \boxed{\textbf{(B) } \sqrt{7}}</math>. ~ciceronii