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

Revision as of 12:53, 16 March 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 1

$[asy] draw((0,0)--(4,5.65)--(8,0)--cycle); label("A", (4,5.65), N, p = fontsize(10pt)); label("C", (8,0), SE, p = fontsize(10pt)); label("B", (0,0), SW, p = fontsize(10pt)); label("P", (3.5,3.5), NW, p = fontsize(10pt)); draw((0,0)--(3.5,3.5)); label("$\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

Solution 2 (Intuition)

$[asy] unitsize(1inch); dot((0.756,0.655)); dot((1.512,1.309)); dot((1.701,0.327)); pair A = origin, B = (1.323,2.291), C = (2.646,0), P = (0.756,0.655), Q = (1.512,1.309), R = (1.701,0.327); draw((0,0)--(1.323,2.291)--(2.646,0)--cycle); label("A", (0,0), SW, p = fontsize(10pt)); label("C", (2.646,0), SE, p = fontsize(10pt)); label("B", (1.323,2.291), N, p = fontsize(10pt)); label("P'", (0.756,0.655), NW, p = fontsize(10pt)); label("1", (2.174,0.164), N, p = fontsize(10pt)); label("1", (1.228,0.491), N, p = fontsize(10pt)); D(A--P, red); D(B--Q, red); D(C--R, red); D(P--Q--R--cycle, blue); D(B--P, magenta); [/asy]$

Suppose that triangle $ABC$ had three segments of length $2$ , emanating from each of its vertices, making equal angles with each of its sides, and going into its interior. Suppose each of these segments intersected the segment clockwise to it precisely at its other endpoint and inside $ABC$ (as pictured in the diagram above). Clearly $s > 2$ and the triangle defined by these intersection points will be equilateral (pictured by the blue segments).

Take this equilateral triangle to have side length $1$ . The portions of each segment outside this triangle (in red) have length $1$ . Take $P'$ to be the intersection of the segments emanating from $A$ and $C$ . By Law of Cosines, $BP' = \sqrt{1 + 1 - 2\cos{120^\circ}} = \sqrt{3}.$ So, $P'$ actually satisfies the conditions of the problem, and we can obtain again by Law of Cosines $s = \sqrt{4 + 1 - 4\cos{120^\circ}} = \boxed{\textbf{(B)} \sqrt{7}}.$

~ hnkevin42

Solution 3 (Heron Bash)

$[asy] draw((0,0)--(4,5.65)--(8,0)--cycle); label("A", (4,5.65), N, p = fontsize(10pt)); label("C", (8,0), SE, p = fontsize(10pt)); label("B", (0,0), SW, p = fontsize(10pt)); label("P", (3.5,3.5), E, p = fontsize(10pt)); label("E", (2.8191,3.982), NW, p = fontsize(10pt)); label("F", (4.848,4.452), NE, p = fontsize(10pt)); label("G", (3.5,0), NE, p = fontsize(10pt)); draw((0,0)--(3.5,3.5)); label("$\sqrt{3}$",(0,0)--(3.5,3.5), SE); draw((8,0)--(3.5,3.5)); label("2",(8,0)--(3.5,3.5), SW); draw((4,5.65)--(3.5,3.5)); label("1",(4,5.65)--(3.5,3.5), E); draw((3.5,3.5)--(2.8191,3.982)); draw((3.5,3.5)--(4.848,4.452)); draw((3.5,3.5)--(3.5,0)); [/asy]$

We begin by dropping altitudes from point $P$ down to all three sides of the triangle as shown above. We can therefore make equations regarding the areas of triangles $\triangle{APC}$ , $\triangle{APB}$ , and $\triangle{BPC}$ . Let $s$ be the side of the equilateral triangle, we use the Heron's formula:

$\triangle{APC} = \frac{s\cdot PF}{2} = \sqrt{\frac{s+3}{2}(\frac{s+3}{2}-s)(\frac{s+3}{2}-1)(\frac{s+3}{2}-2)}$ $\implies PF = \frac{\sqrt{10s^2-4s-9}}{2s}$

Similarly, we obtain:

$PE = \frac{\sqrt{-s^4+8s^2-4}}{2s}$ $PG = \frac{\sqrt{-s^4+14s^2-1}}{2s}$

By Viviani's theorem, $\frac{\sqrt{10s^2-4s-9}}{2s}+\frac{\sqrt{-s^4+8s^2-4}}{2s}+\frac{\sqrt{-s^4+14s^2-1}}{2s} = \frac{\sqrt{3}}{2}s$ $\sqrt{10s^2-4s^4-9}+\sqrt{-s^4+8s^2-4}+\sqrt{-s^4+14s^2-1} = \sqrt{3}s^2$

Video Solution

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

@@ Line 105: / Line 105: @@
 By Viviani's theorem,
 <cmath>\frac{\sqrt{10s^2-4s-9}}{2s}+\frac{\sqrt{-s^4+8s^2-4}}{2s}+\frac{\sqrt{-s^4+14s^2-1}}{2s} = \frac{\sqrt{3}}{2}s</cmath>
-<cmath>\sqrt{10s^2-4s-9}+\sqrt{-s^4+8s^2-4}+\sqrt{-s^4+14s^2-1} = \sqrt{3}s^2</cmath>
+<cmath>\sqrt{10s^2-4s^4-9}+\sqrt{-s^4+8s^2-4}+\sqrt{-s^4+14s^2-1} = \sqrt{3}s^2</cmath>
 ==Video Solution==

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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