Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Line $l$ in the coordinate plane has equation $3x-5y+40=0$ . This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$ . What is the $x$ -coordinate of the $x$ -intercept of line $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$

Solution: A quick check tells us that $(20,20)$ falls on the given line. Common sense tells us that if the slope of the original line is $1$ , or 45 degrees from the horizontal, a 45 counter clockwise rotation will result in a vertical line, and the x-intercept will be $(20,0)$ . 
Thinking of a 45 degree counter clockwise rotation as a 90 degree counter clockwise rotation that is bisected will helps in visualizing this line of reasoning.
Therefore, it follows that if the original line is made steeper, then the x-intercept will move away from $(20,0)$ to the right. If the original line is made lower, then the opposite will happen. Our given line has slope $3/5$ , so the answer must be $A$ or $B$ $A$ can be eliminated because an x-intercept of $(10,0)$ can only occur when the original line is horizontal. In conclusion, the answer must be $\boxed{15}$

Answer: B 15
Problem: There are integers $a, b,$ and $c,$ each greater than $1,$ such that
\[\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}\]
for all $N \neq 1$ . What is $b$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$

Solution: $\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}}$ can be simplified to $N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}.$
The equation is then $N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}=N^{\frac{25}{36}}$ which implies that $\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.$
$a$ has to be $2$ since $\frac{25}{36}>\frac{7}{12}$ $\frac{7}{12}$ is the result when $a, b,$ and $c$ are $3, 2,$ and $2$
$b$ being $3$ will make the fraction $\frac{2}{3}$ which is close to $\frac{25}{36}$
Finally, with $c$ being $6$ , the fraction becomes $\frac{25}{36}$ . In this case $a, b,$ and $c$ work, which means that $b$ must equal $\boxed{3.}$ ~lopkiloinm

Answer: B 3.
Problem: There are integers $a, b,$ and $c,$ each greater than $1,$ such that
\[\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}\]
for all $N \neq 1$ . What is $b$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$

Solution: As above, notice that you get $\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.$
Now, combine the fractions to get $\frac{bc+c+1}{abc}=\frac{25}{36}$
Let us assume $bc+c+1=25$ and $abc=36$ as this is the most convenient. (EDIT: This used to say WLOG but that is inaccurate)
From the first equation, we get $c(b+1)=24$ . Note also that from the second equation, $b$ and $c$ must be factors of 36.
After listing out the factors of 36 and utilising trial and error, we find that $c=6$ and $b=3$ works, with $a=2$ . So our answer is $\boxed{3.}$

Answer: B 3.
Problem: There are integers $a, b,$ and $c,$ each greater than $1,$ such that
\[\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}\]
for all $N \neq 1$ . What is $b$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$

Solution: Collapsed, $\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[abc]{N^{bc+c+1}}$ . Comparing this to $\sqrt[36]{N^{25}}$ , observe that $bc+c+1=25$ and $abc=36$ . The first can be rewritten as $c(b+1)=24$ . Then, $b+1$ has to factor into 24 while 1 less than that also must factor into 36. The prime factorizations are as follows $36=2^2 3^2$ and $24=2^33$ . Then, $b=\boxed{3}$ , as only 4 and 3 factor into 36 and 24 while being 1 apart.

Answer: B 3
Problem: The vertices of a quadrilateral lie on the graph of $y=\ln{x}$ , and the $x$ -coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln{\frac{91}{90}}$ . What is the $x$ -coordinate of the leftmost vertex?
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 13$

Solution: Let the coordinates of the quadrilateral be $(n,\ln(n)),(n+1,\ln(n+1)),(n+2,\ln(n+2)),(n+3,\ln(n+3))$ . We have by shoelace's theorem, that the area is \begin{align*} &\frac{\ln(n)(n+1) + \ln(n+1)(n+2) + \ln(n+2)(n+3)+n\ln(n+3)}{2} - \frac{\ln(n+1)(n) + \ln(n+2)(n+1) + \ln(n+3)(n+2)+\ln(n)(n+3)}{2} \\ &=\frac{\ln \left( \frac{n^{n+1}(n+1)^{n+2}(n+2)^{n+3}(n+3)^n}{(n+1)^n(n+2)^{n+1}(n+3)^{n+2}n^{n+3}}\right)}{2} \\ &= \ln \left( \sqrt{\frac{(n+1)^2(n+2)^2}{n^2(n+3)^2}} \right)  \\ &= \ln \left(\frac{(n+1)(n+2)}{n(n+3)}\right) \\ &= \ln \left( \frac{91}{90} \right). \end{align*} We know that the numerator must have a factor of $13$ , so given the answer choices, $n$ is either $12$ or $11$ . If $n=11$ , the expression $\frac{(n+1)(n+2)}{n(n+3)}$ does not evaluate to $\frac{91}{90}$ , but if $n=12$ , the expression evaluates to $\frac{91}{90}$ . Hence, our answer is $\boxed{12}$

Answer:  12
Problem: The vertices of a quadrilateral lie on the graph of $y=\ln{x}$ , and the $x$ -coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln{\frac{91}{90}}$ . What is the $x$ -coordinate of the leftmost vertex?
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 13$

Solution: Like above, use the shoelace formula to find that the area of the quadrilateral is equal to $\ln\frac{(n+1)(n+2)}{n(n+3)}$ . Because the final area we are looking for is $\ln\frac{91}{90}$ , the numerator factors into $13$ and $7$ , which one of $n+1$ and $n+2$ has to be a multiple of $13$ and the other has to be a multiple of $7$ . Clearly, the only choice for that is $\boxed{12}$

Answer:  12
Problem: How many positive integers $n$ are there such that $n$ is a multiple of $5$ , and the least common multiple of $5!$ and $n$ equals $5$ times the greatest common divisor of $10!$ and $n?$
$\textbf{(A) } 12 \qquad \textbf{(B) } 24 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 72$

Solution: We set up the following equation as the problem states:
\[\text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}.\]
Breaking each number into its prime factorization, we see that the equation becomes
\[\text{lcm}{(2^3\cdot 3 \cdot 5, n)} = 5\text{gcd}{(2^8\cdot 3^4 \cdot 5^2 \cdot 7, n)}.\]
We can now determine the prime factorization of $n$ . We know that its prime factors belong to the set $\{2, 3, 5, 7\}$ , as no factor of $10!$ has $11$ in its prime factorization, nor anything greater. Next, we must find exactly how many different possibilities exist for each.
There can be anywhere between $3$ and $8$ $2$ 's and $1$ to $4$ $3$ 's. However, since $n$ is a multiple of $5$ , and we multiply the $\text{gcd}$ by $5$ , there can only be $3$ $5$ 's in $n$ 's prime factorization. Finally, there can either $0$ or $1$ $7$ 's.
Thus, we can multiply the total possibilities of $n$ 's factorization to determine the number of integers $n$ which satisfy the equation, giving us $6 \times 4 \times 1 \times 2 = \boxed{48}$ . ~ciceronii

Answer: D 48
Problem: How many positive integers $n$ are there such that $n$ is a multiple of $5$ , and the least common multiple of $5!$ and $n$ equals $5$ times the greatest common divisor of $10!$ and $n?$
$\textbf{(A) } 12 \qquad \textbf{(B) } 24 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 72$

Solution: Like the Solution 1, we start from the equation:
\[\text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}.\] Assume $\text{lcm}{(5!, n)}=k\cdot5!$ , with some integer $k$ . It follows that $k\cdot 4!=\text{gcd}{(10!, n)}$ . It means that $n$ has a divisor $4!$ . Since $n$ is a multiple of $5$ $n$ has a divisor $5!$ . Thus, $\text{lcm}{(5!, n)}=n=k\cdot5!$ . The equation can be changed as \[k\cdot5!=5\text{gcd}{(10!, k\cdot5!)}\] \[k=5\text{gcd}{(6\cdot7\cdot8\cdot9\cdot10, k)}\] We can see that $k$ is also a multiple of $5$ , with a form of $5\cdot m$ . Substituting it in the above equation, we have \[m=5\text{gcd}{(6\cdot7\cdot8\cdot9\cdot2, m)}\] Similarly, $m$ is a multiple of $5$ , with a form of $5\cdot s$ . We have \[s=\text{gcd}{(6\cdot7\cdot8\cdot9\cdot2, 5\cdot s)}=\text{gcd}{(2^5\cdot3^3\cdot7, s)}\] The equation holds, if $s$ is a divisor of $2^5\cdot3^3\cdot7$ , which has $(5+1)(3+1)(1+1)=\boxed{48}$ divisors. ~Linty Huang

Answer: D 48
Problem: How many positive integers $n$ are there such that $n$ is a multiple of $5$ , and the least common multiple of $5!$ and $n$ equals $5$ times the greatest common divisor of $10!$ and $n?$
$\textbf{(A) } 12 \qquad \textbf{(B) } 24 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 72$

Solution: As in the previous solutions, we start with
\[\text{lcm}(5!,n) = 5\text{gcd}(10!,n)\]
From this we have that $\text{lcm}(5!,n) \,|\, 5\text{gcd}(10!,n)$ , and in particular, $n \,|\, 5\text{gcd}(10!,n)$ . However, $\text{gcd}(10!,n)\, |\, n$ , so we must have $\text{gcd}(10!,n) = n$ or $\text{gcd}(10!,n) = n/5$ . If $\text{gcd}(10!,n) = n$ , then we have $\text{lcm}(5!,n) = 5n$ ; because $5\, |\, 5!$ , this implies that 5 does not divide $n$ , so we must have $\text{gcd}(10!,n) = n/5$
Now we have $\text{lcm}(5!,n) = n$ , implying that $5!\, |\, n$ , and $n/5\, |\, 10!$ . Writing out prime factorizations, this gives us
\[2^3 \cdot 3 \cdot 5 \,|\, n\] \[n \,|\, 2^8 \cdot 3^4 \cdot 5^2 \cdot 7\]
So $n$ can have 3, 4, 5, 6, 7, or 8 factors of 2; 1, 2, 3, or 4 factors of two; and 0 or 1 factors of 7. Note that $\text{gcd}(2^8 \cdot 3^4 \cdot 5^2 \cdot 7,n) = n/5$ implies that $n$ has 2 factors of 5. Thus, there are $6 \cdot 4 \cdot 2 = 48$ possible choices for $n$ , and our answer is $\boxed{48}$

Answer: D 48
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: 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 $s = \boxed{7}$

Answer: B 7
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: 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 $\boxed{7}$

Answer: B 7
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: 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 $\boxed{7}$

Answer: B 7
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: [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 \[s = \sqrt{4 + 1 - 4\cos{120^\circ}} = \boxed{7}.\]

Answer: B 7
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: [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 $\boxed{7}$

Answer: B 7
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: 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 $\boxed{7}$

Answer: B 7
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: 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 $\boxed{7}$

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

Answer: B 7