2020 AMC 12A Problems/Problem 24

Problem

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

$[asy] unitsize(0.5inch); pen p = fontsize(10pt); draw((0,0)--(4,5.65)--(8,0)--cycle); label("$A$", (4,5.65), N, p); label("$C$", (8,0), SE, p); label("$B$", (0,0), SW, p); label("$P$", (3.5,3.5), NW, p); draw((0,0)--(3.5,3.5)); label("$\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); 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 $60^{\circ}$ counterclockwise 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: $\begin{align*} s^2 &= (AP)^2 + (CP)^2 - 2(AP)(CP)\cos{\angle{APC}} \\ &= 1 + 4 - 2(1)(2)\cos{120^{\circ}} \\ &= 5 - 4\left(-\frac{1}{2}\right) \\ &= 7, \end{align*}$ giving us that $s = \boxed{\textbf{(B) } \sqrt{7}}$ .

~ciceronii

Solution 2 (Intuition)

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

$[asy] draw((0,0)--(4,5.65)--(8,0)--cycle); label("$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$", (3.5,0), down, p = fontsize(10pt)); draw((0,0)--(3.5,3.5)); label("$\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:

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

Similarly, we obtain:

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

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

Note that from now on, the algebra will get extremely ugly and almost impossible to do by hand within the time frame. However, we do see that it's extremely easy to check the answer choices with the equation in this form. Testing $s = \sqrt{7}$ , We obtain $7\sqrt{3}$ on both sides, revealing that our answer is in fact $\boxed{\textbf{(B) } \sqrt{7}}$

~ siluweston ~ edits by aopspandy

Solution 4 (Area)

Instead of directly finding the side length of the equilateral triangle, we instead find the area and use it to find the side length. Begin by reflecting $P$ over each of the sides. Label these reflected points $P', P'', P'''$ . Connect these points to the vertices of the equilateral triangle, as well as to each other.

$[asy] draw((0,3.5)--(4,9.15)--(8,3.5)--cycle); label("$A$", (4,9.15), N, p = fontsize(10pt)); label("$C$", (8,3.5), SE, p = fontsize(10pt)); label("$B$", (0,3.5), SW, p = fontsize(10pt)); label("$P$", (3.5,7), NW, p = fontsize(10pt)); draw((0,3.5)--(3.5,7)); draw((8,3.5)--(3.5,7)); draw((4,9.15)--(3.5,7)); label("$P'$", (3.5, 0), S, p = fontsize(10pt)); draw((8,3.5)--(3.5,0)); draw((0,3.5)--(3.5,0)); label("$P''$",(6,8.5), NE, p = fontsize(10pt)); draw((4,9.15)--(6,8.5)); draw((8,3.5)--(6,8.5)); label("$P'''$",(2.25,8), NW, p = fontsize(10pt)); draw((0,3.5)--(2.25,8)); draw((4,9.15)--(2.25,8)); draw((3.5,0)--(6,8.5)--(2.25,8)--cycle); [/asy]$

Observe that the area of the equilateral triangle $ABC$ is half that of the hexagon $AP''CP'BP'''$ .

Note that $AP=AP''=AP'''$ . The same goes for the other vertices. This means that $AP''P'''$ is isosceles. Using either the Law of Cosines or simply observing that $AP''P'''$ is comprised of two 30-60-90 triangles, we find that $P''P'''= \sqrt{3}$ . Similarly (pun intended), $P'P'''=3$ and $P'P''=2\sqrt{3}$ . Using the previous observation that $AP''P'''$ is two 30-60-90 triangles (as are the others) we find the areas of $AP''P''$ to be $\frac{\sqrt{3}}{4}$ . Again, using similarity we find the area of $BP'P'''$ to be $\frac{3\sqrt{3}}{4}$ and the area of $CP'P''$ to be $\sqrt{3}$ .

Next, observe that $P'P''P'''$ is a 30-60-90 right triangle. This right triangle therefore has an area of $\frac{3\sqrt{3}}{2}$ .

Adding these areas together, we get the area of the hexagon as $\frac{7\sqrt{3}}{2}$ . This means that the area of $ABC$ is $\frac{7\sqrt{3}}{4}$ .

The formula for the area of an equilateral triangle with side length $s$ is $\frac{s^2\sqrt{3}}{4}$ (if you don't have this memorized it's not hard to derive). Comparing this formula to the area of $ABC$ , we can easily find that $s^2=7$ , which means that the side length of $ABC$ is $\boxed{\textbf{(B) } \sqrt{7}}$ .

While this approach feels rather convoluted in comparison to Solution 1, it is more flexible and can actually be generalized for any point in an equilateral triangle (although that requires use of Heron's).

~IAmTheHazard

Solution 5

Suppose $A(0,\sqrt{3}a)$ , $B(-a,0)$ , $C(a,0)$ and $P(x,y)$ . So $s=2a$ . Since $BP = \sqrt{3}$ and $CP = 2$ , we have $(x+a)^2+y^2=3$ $(x-a)^2+y^2=4$ Solving the equations, we have $x=-\frac{1}{4a},~~y=\sqrt{\frac72-a^2-\frac{1}{16a^2}}$ From $AP=1$ , we can have $a=\sqrt{7}/2$ . The answer is $\boxed{\textbf{(B) } \sqrt{7}}$ .

~Linty Huang

Solution 6

Rotate $\triangle CPA$ countclockwise $60^\circ$ along point $C$ to $\triangle CQB$ . Then $CP=CQ, \angle PCQ=60^\circ$ , so $\triangle CPQ$ is an equilateral triangle. Note that $\triangle PQB$ is a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle, hence $\angle BPQ=30^\circ$ , and $\angle BPC=90^\circ$ , $BC^2=PC^2+PB^2=2^2+3=7$ , the answer is $\boxed{\textbf{(B) } \sqrt{7}}$ . $[asy] size(7cm); draw((0,0) -- (2.7,0)); draw((0,0) -- (1.35,2.34)); draw((2.7,0)--(1.35, 2.34)); label("$A$", (0,0), W); label("$B$", (2.7,0), E); label("$C$", (1.35, 2.34), N); draw((0,0)-- (0.92, 0.39)); draw((2.7,0)-- (0.92, 0.39)); draw((1.35, 2.34)-- (0.92, 0.39)); draw((1.35, 2.34)-- (2.83, 0.99), dashed); draw((2.7,0)-- (2.83, 0.99), dashed); draw((0.92,0.39)-- (2.83, 0.99), dashed); label("$P$", (0.92,0.39), NW); label("$Q$", (2.83,0.99), NE); [/asy]$ ~szhang

Solution 7 (Most elegant of them all)

Drawing out a rough sketch, it appears that $\angle BPC = 90^{\circ}$ . By Pythagorean, our answer is $\sqrt{\sqrt{3}^2 + 2^2} = \boxed{\textbf{(B) } \sqrt{7}}$ .

Video Solutions

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

Video Solution by Richard Rusczyk - https://www.youtube.com/watch?v=xnAXGUthO54&list=PLyhPcpM8aMvJvwA2kypmfdtlxH90ShZCc&index=4 - AMBRIGGS

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

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