2020 AMC 12A Problems/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 (Rotations)

Solution 1(a)

We begin by rotating $\triangle{ APB}$ counterclockwise by $60^{\circ}$ about $A$, such that $P\mapsto Q$ and $B\mapsto C$. We see that $\triangle{ APQ}$ is equilateral with side length $1$, meaning that $\angle APQ = 60^{\circ}$. We also see that $\triangle{CPQ}$ is a $30$-$60$-$90$ right triangle, meaning that $\angle CPQ= 60^{\circ}$. Thus, by adding the two together, we see that $\angle APC = 120^{\circ}$. [asy] size(200); pen p = fontsize(10pt)+gray+0.5; pen q = fontsize(13pt); pair A,B,C,D,P,Q; real s=sqrt(7); B=origin; A=s*dir(60); C=s*right; P=IP(CR(A,1),CR(C,2)); Q=rotate(60,A)*P; draw(A--B--C--A, black+0.8); draw(A--P--B^^P--C^^A--Q--C, p); draw(P--Q, p+dashed); //draw(A--A+C--C, p); label("$A$", A, up, q);  label("$B$", B, 0.5*(B-P), q); label("$C$", C, 0.5*(C-P), q);   label("$P$", P, dir(180), q);   label("$Q$", Q, 0.25*(Q-B), q);  label("$\sqrt{3}$",B--P, right, p);  label("$2$",C--P, 2*left, p);  label("$1$",A--P, 1.5**dir(-10), p);   label("$1$", A--Q, dir(250), p);  label("$1$",P--Q, down, p);  label("$\sqrt{3}$",C--Q, right, p);  [/asy] We can now use the law of cosines as following: \begin{align*} s^2 &= (AP)^2 + (CP)^2 - 2\cdot AP\cdot CP\cdot \cos{\angle{APC}} \\ &= 1 + 4 - 2\cdot 1\cdot 2\cdot \cos{120^{\circ}} \\ &= 5 - 4\left(-\frac{1}{2}\right) \\ &= 7, \end{align*} giving us that $s = \boxed{ \sqrt{7} \ \textbf{(B)}}$.


Solution 1(b)

Rotate $\triangle CPA$ counterclockwise $60^\circ$ around point $C$ to $\triangle CQB$. Then $CP=CQ, \angle PCQ=60^\circ$, so $\triangle CPQ$ is an equilateral triangle. [asy] size(200); pen p = fontsize(10pt)+gray+0.4; pen q = fontsize(13pt); pair A,B,C,D,P,Q; real s=sqrt(7); A=origin; B=s*right; C=s*dir(60); P=IP(CR(A,1),CR(C,2)); Q=rotate(60,C)*P; draw(A--B--C--A, black+0.8); draw(A--P--B^^P--C, p); draw(B--Q--C^^P--Q, p+dashed); label("$A$", A, A-P, q);  label("$B$", B, B-P, q); label("$C$", C, up, q);   label("$P$", P, dir(140), q);   label("$Q$", Q, 0.25*(Q-P), q);  label("$\sqrt{3}$",P--B, dir(60), p);  label("$2$",C--P, 2*left, p);  label("$1$",A--P, up, p);   label("$1$", B--Q, dir(-30), p);  label("$2$",P--Q, down, p);  label("$2$",C--Q, dir(45), p);  [/asy] Note that $\triangle PQB$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, hence $\angle BPQ=30^\circ$, and $\angle BPC=90^\circ$, so \[BC^2=PC^2+PB^2=2^2+3=7,\] and the answer is $\boxed{\textbf{(B) } \sqrt{7}}$.


Solution 1(c)

Rotate $\triangle BPC$ counterclockwise by $60^\circ$ around point $B$ to $\triangle BQA$. Then $BP=BQ$, and $\angle PBQ=60^\circ$, so $\triangle BPQ$ is an equilateral triangle. [asy] size(200); pen p = fontsize(10pt)+gray+0.4; pen q = fontsize(13pt); pair A,B,C,D,P,Q,R,X,Y,Z; real s=sqrt(7); C=origin; A=s*right; B=s*dir(60); P=IP(CR(A,1),CR(C,2)); Q=rotate(60,B)*P;  draw(A--B--C--A, black+0.8); draw(A--P--B^^P--C, p); draw(B--Q--A^^P--Q, p+dashed); label("$A$", A, A-P, q);  label("$B$", B, B-P, q); label("$C$", C, 0.6*(C-P), q);   label("$P$", P, dir(250), q);   label("$Q$", Q, 0.25*(Q-P), q);  label("$\sqrt{3}$",B--Q, dir(60), p); label("$\sqrt{3}$",P--B, dir(210), p); label("$\sqrt{3}$",P--Q, dir(135), p); label("$2$",P--C, dir(120), p); label("$2$",A--Q, dir(-20), p); label("$1$",A--P, dir(210), p); [/asy] Note that $\triangle QAP$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, hence $\angle PQA=30^\circ$, and $\angle BQA=90^\circ$, so \[AB^2=PQ^2+AQ^2=2^2+3=7,\] and the answer is $\boxed{\textbf{(B) } \sqrt{7}}$.

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] unitsize(0.4inch); 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), E, p); label("$E$", (2.8191,3.982), NW, p); label("$F$", (4.848,4.452), NE, p); label("$G$", (3.5,0), down, p); draw((0,0)--(3.5,3.5)); label("$\sqrt{3}$",(0,0)--(3.5,3.5), SE,p); draw((8,0)--(3.5,3.5)); label("$2$",(8,0)--(3.5,3.5), SW,p); draw((4,5.65)--(3.5,3.5)); label("$1$",(4,5.65)--(3.5,3.5), E,p); 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] size(300); 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 (which only works for isosceles triangles), it is more flexible and can actually be generalized for any point in a general triangle (although that requires use of Heron's, and potentially Law of Sines and Cosines).


Solution 5 (Coordinate Bashing)

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 $AP=1$ (and a fair amount of algebra), we can have $a=\sqrt{7}/2$. The answer is $\boxed{\textbf{(B) } \sqrt{7}}$.

~Linty Huang

Solution 6 (Diagram Nuke)

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

Proof of this fact can be found in the Video Solution by Richard Rusczyk below.

Solution 7 (Theorem Nuke)

We can use the following theorem:

$\text{For a point } P \text{ inside of an equilateral triangle } \triangle ABC \text{ and side length } s$, \[3(PA^4+PB^4+PC^4 + s^4)=(PA^2+PB^2+PC^2+s^2)^2\]

We know that $PA=1, PB=\sqrt{3}$, and $CP=2$. Plugging these into our formula, we get $3(1+9+16+s^4)=(1+3+4+s^2)^2 \Rightarrow 78 + 3s^4=64+16s^2+s^4 \Rightarrow 2s^4-16s^2+14=0$. Let $s^2=u$. Then, we have $2u^2-16u+14=0$. Solving for $u$, we get $\frac{16 \pm \sqrt{144}}{4} = 4 \pm 3$. If $u$ is equal to $1$, then we have $s=1$, but this is not possible since $P$ is inside of the triangle. This means that $u=7$, and therefore $s=\boxed{\textbf{(B) } \sqrt{7}}$.


Solution 8 (More Coordinate Bashing)

Set the points $A(0,0), B\left(\frac{s}{2},\frac{s\sqrt{3}}{2}\right), C(s,0)$. Then, we want to find the intersection of the three circles \[x^2+y^2=1, (x-s)^2+y^2=4, \left(x-\frac{s}{2}\right)^2+\left(y-\frac{s\sqrt{3}}{2}\right)^2=3.\] Subtracting circle 1 from circle 2 yields $s^2-2xs=3$, which can be rewritten as $x=\frac{s^2-3}{2s}$. Subtracting circle 1 from circle 3 yields $s^2-sx-ys\sqrt{3}=2$, or $2s^2-2sx-2ys\sqrt{3}=4$. Subtracting $s^2-2xs=3$ from this yields $s^2-2ys\sqrt{3}=1$, or $y=\frac{s^2-1}{2s\sqrt{3}}$.

We then substitute our values for $x$ and $y$ back into $x^2+y^2=1$, which gives us \[\left(\frac{s^2-3}{2s}\right)^2+\left(\frac{s^2-1}{2s\sqrt{3}}\right)^2=1.\] Solving this equation (the algebra is surprisingly not bad!) gives us $s=\pm1,\pm\sqrt{7}$. Since $s$ must be greater than 1, the answer is $\boxed{\textbf{(B) } \sqrt{7}}$.

- curiousmind888 & TGSN

Solution 9 (Trig Bash)

Let $\angle APC = \alpha$, $\angle CPB = \beta$, by the Law of Cosine \[1^2+2^2 - 2 \cdot 1 \cdot 2 \cdot \cos \alpha = 2^2 + (\sqrt{3})^2 - 2 \cdot 2 \cdot \sqrt{3} \cdot \cos \beta\] \[2^2 + (\sqrt{3})^2 - 2 \cdot 2 \cdot \sqrt{3} = (\sqrt{3})^2 + 1^2 - 2 \cdot \sqrt{3} \cdot 1 \cdot \cos (360^{\circ} - \alpha - \beta)\] Simplifying the two equations we get \[\sqrt{3} \cos \beta - \cos \alpha = \frac12\] \[3 - 4 \sqrt{3} \cos \beta = - 2 \sqrt{3} ( \cos \alpha \cos \beta - \sin \alpha \sin \beta )\]

Let $\cos \alpha = x$, $\cos \beta = y$, by simplifying we get: \[y\sqrt{3} - x = \frac12\] \[3-4y\sqrt{3} = -2\sqrt{3} \left( xy - \sqrt{ (1-x^2)(1-y^2) } \right)\] Solving the first equation for $y$ we get $y = \frac{ 2x \sqrt{3} + \sqrt{3} }{6}$

Substituting this into the second equation gives \[3-4\sqrt{3} \cdot \frac{ 2x \sqrt{3} + \sqrt{3} }{6} = -2\sqrt{3} \left( x \cdot \frac{ 2x \sqrt{3} + \sqrt{3} }{6} - \sqrt{ (1-x^2)(1- (\frac{ 2x \sqrt{3} + \sqrt{3} }{6})^2 ) } \right)\] \[3-2(2x+1) = - 2\sqrt{3} \cdot x \cdot \frac{ 2x \sqrt{3} + \sqrt{3} }{6} + 2\sqrt{3} \cdot \sqrt{ (1-x^2)(1- (\frac{ 2x \sqrt{3} + \sqrt{3} }{6})^2 ) }\] \[3 - 2x - 2 = -2x^2 - x + 2\sqrt{3} \cdot \sqrt{ (1-x^2)(1- (\sqrt{3})^2(\frac{ 2x + 1 }{6})^2 ) }\] \[3 - 2x - 2 + 2x^2 + x = \sqrt{ 12 (1-x^2)(1- \frac{ 4x^2 + 4x + 1 }{12} ) }\] \[1 - 3x + 2x^2 = \sqrt{ (1-x^2)(11 - 4x^2 - 4x) }\] \[(1 - 3x + 2x^2)^2 = (1-x^2)(11 - 4x^2 - 4x)\] \[1 - 3x + 2x^2 - 3x + 9x^2 - 6x^3 + 2x^2 - 6x^3 + 4x^4 = 11 - 4x - 4x^2 - 11x^2 + 4x^3 + 4x^4\] \[1 - 6x + 13x^2 - 12x^3 = 11 - 4x - 15x^2 + 4x^3\] \[8x^3 - 14x^2 + x + 5 = 0\] \[(x-1)(4x-5)(2x+1) = 0\] \[\cos \alpha<1, \quad \cos \alpha = x = -\frac12, \quad s = \sqrt{ 5 - 4 \cos \alpha } = \boxed{\textbf{(B) } \sqrt{7}}\]


Video Solutions


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

See Also

2020 AMC 12A (ProblemsAnswer KeyResources)
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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