Answer: 25.0

Let $\mathcal{T}$ be the set of ordered triples $(x,y,z)$ of nonnegative real numbers that lie in the plane $x+y+z=1.$ Let us say that $(x,y,z)$ supports $(a,b,c)$ when exactly two of the following are true: $x\ge a, y\ge b, z\ge c.$ Let $\mathcal{S}$ consist of those triples in $\mathcal{T}$ that support $\left(\frac 12,\frac 13,\frac 16\right).$ The area of $\mathcal{S}$ divided by the area of $\mathcal{T}$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$

Solution:
This problem just requires a good diagram and strong 3D visualization.
1999 AIME-8.png
The region in $(x,y,z)$ where $x \ge \frac{1}{2}, y \ge \frac{1}{3}$ is that of a little triangle on the bottom of the above diagram, of $y \ge \frac{1}{3}, z \ge \frac{1}{6}$ is the triangle at the right, and $x \ge \frac 12, z \ge \frac 16$ the triangle on the left, where the triangles are coplanar with the large equilateral triangle formed by $x+y+z=1,\ x,y,z \ge 0$ . We can check that each of the three regions mentioned fall under exactly two of the inequalities and not the third.
1999 AIME-8a.png
The side length of the large equilateral triangle is $\sqrt{2}$ , which we can find using 45-45-90 $\triangle$ with the axes. Using the formula $A = \frac{s^2\sqrt{3}}{4}$ for equilateral triangles , the area of the large triangle is $\frac{(\sqrt{2})^2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$ . Since the lines of the smaller triangles are parallel to those of the large triangle, by corresponding angles we see that all of the triangles are similar , so they are all equilateral triangles. We can solve for their side lengths easily by subtraction, and we get $\frac{\sqrt{2}}{6}, \frac{\sqrt{2}}{3}, \frac{\sqrt{2}}{2}$ . Calculating their areas, we get $\frac{\sqrt{3}}{8}, \frac{\sqrt{3}}{18}, \frac{\sqrt{3}}{72}$ . The ratio $\frac{\mathcal{S}}{\mathcal{T}} = \frac{\frac{9\sqrt{3} + 4\sqrt{3} + \sqrt{3}}{72}}{\frac{\sqrt{3}}{2}} = \frac{14}{36} = \frac{7}{18}$ , and the answer is $m + n = \boxed{025}$


Answer: 276.0

Let $\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$ , where $0<r$ and $0\leq \theta <360$ . Find $\theta$

Solution:
\begin{eqnarray*} 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \frac{z^5-1}{z-1}\\ 0 &=& \frac{(z^5 - 1)(z(z-1)+1)}{z-1} = \frac{(z^2-z+1)(z^5-1)}{z-1} \end{eqnarray*}
Thus $z^5 = 1, z \neq 1 \Longrightarrow z = \mathrm{cis}\ 72, 144, 216, 288$
or $z^2 - z + 1 = 0 \Longrightarrow z = \frac{1 \pm \sqrt{-3}}{2} = \mathrm{cis}\ 60, 300$
(see cis ).
Discarding the roots with negative imaginary parts (leaving us with $\mathrm{cis} \theta,\ 0 < \theta < 180$ ), we are left with $\mathrm{cis}\ 60, 72, 144$ ; their product is $P = \mathrm{cis} (60 + 72 + 144) = \mathrm{cis} \boxed{276}$

Solution: Divide through by $z^3$ . We get the equation $z^3 + \frac {1}{z^3} + z + \frac {1}{z} + 1 = 0$ . Let $x = z + \frac {1}{z}$ . Then $z^3 + \frac {1}{z^3} = x^3 - 3x$ . Our equation is then $x^3 - 3x + x + 1 = x^3 - 2x + 1 = (x - 1)(x^2 + x - 1) = 0$ , with solutions $x = 1, \frac { - 1\pm\sqrt {5}}{2}$ . For $x = 1$ , we get $z = \text{cis}60,\text{cis}300$ . For $x = \frac { - 1 + \sqrt {5}}{2}$ , we get $z = \text{cis}{72},\text{cis}{292}$ (using exponential form of $\cos$ ). For $x = \frac { - 1 - \sqrt {5}}{2}$ , we get $z = \text{cis}144,\text{cis}216$ . The ones with positive imaginary parts are ones where $0\le\theta\le180$ , so we have $60 + 72 + 144 = \boxed{276}$

Solution:
We recognize that $z^6+z^5+z^4+z^3+z^2+z+1=\frac{z^7-1}{z-1}$ and strongly wish for the equation to turn into so. Thankfully, we can do this by simultaneously adding and subtracting $z^5$ and $z$ as shown below. $z^6+z^5+z^4+z^3+z^2+z+1-(z^5+z)=0\implies \frac{z^7-1}{z-1}-(z^5+z)=0$
Now, knowing that $z=1$ is not a root, we multiply by $z-1$ to obtain $z^7-1-(z-1)(z^5+z)=0\implies z^7-1-(z^6+z^2-z^5-z)=0\implies z^7-z^6+z^5-z^2+z-1=0$ Now, we see the $z^2+z-1$ and it reminds us of the sum of two cubes. Cleverly factoring, we obtain that.. $z^5(z^2-z)+z^5-(z^2-z+1)=0\implies z^5(z^2-z+1)-(z^2-z+1)=0\implies (z^5-1)(z^2-z+1)=0$
Now, it is clear that we have two cases to consider.
Case $1$ $z^5-1=0$ We obtain that $z^5=1$ or $z^5=e^{2\pi{n}{i}}$ Obviously, the answers to this case are $e^{ia}, a\in{\frac{2\pi}{5}, \frac{4\pi}{5}}$
Case $2$ $z^2-z+1=0$ Completing the square and then algebra allows us to find that $z=\frac{1}{2}\pm\frac{i\sqrt{3}}{2}$ which has $\arg$ $\frac{\pi}{3}$
Hence, the answer is $\frac{18\pi}{15}+\frac{5\pi}{15}=\frac{23\pi}{15}\cdot\frac{180}{\pi}=\boxed{276}$

Solution:
Add 1 to both sides of the equation to get $x^6+x^4+x^3+x^2+1+1=1$ . We can rearrange to find that $(x^6+x^3+1)+(x^4+x^2+1)=1$ . Then, using sum of a geometric series, $\frac{x^9-1}{x^3-1}+\frac{x^6-1}{x^2-1}=1$
Combining the two terms of the LHS, we get that $\frac{x^{11}-x^9-x^2+1+x^9-x^6-x^3+1}{x^5-x^3-x^2+1}=1$ , so $x^{11}-x^6-x^3-x^2+2=x^5-x^3-x^2+1$ , and simplifying, we see that $x^{11}-x^6-x^5+1=0$ , so by SFFT, $(x^6-1)(x^5-1)=0$ . Then, the roots of our polynomial are the fifth and sixth roots of unity. However, looking back at our expression when it had fractions, we realize that if $x$ is a second or third root of unity, it would cause a denominator to be zero, so the roots of our polynomial are the fifth and sixth roots of unity that are not the second or third roots of unity. The only roots in this category are $\mathrm{cis} (60, 72, 144)$ , so our desired sum is $\boxed{276}$ , and we are done.


Answer: 24.0

Let $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ where $i = \sqrt{-1}.$ Find the value of the product \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right).\]

Solution:
For any $k\in Z$ , we have, \begin{align*} & \left( \omega^{3k} + \omega^k + 1 \right) \left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\ & = \omega^{3 \cdot 7} + \omega^{2k + 7} + \omega^{3k} + \omega^{-2k + 3 \cdot 7} + \omega^7 + \omega^k + \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \\ & = 1 + \omega^{2k} + \omega^{3k} + \omega^{-2k} + 1 + \omega^k + \omega^{-3k} + \omega^{-k} + 1 \\ & = 2 + \omega^{-3k} \sum_{j=0}^6 \omega^{j k} \\ & = 2 + \omega^{-3k} \frac{1 - \omega^{7 k}}{1 - \omega^k} \\ & = 2 . \end{align*} The second and the fifth equalities follow from the property that $\omega^7 = 1$
Therefore, \begin{align*} \Pi_{k=0}^6 \left( \omega^{3k} + \omega^k + 1 \right) & = \left( \omega^{3 \cdot 0} + \omega^0 + 1 \right) \Pi_{k=1}^3 \left( \omega^{3k} + \omega^k + 1 \right) \left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\ & = 3 \cdot 2^3 \\ & = \boxed{024}

Solution:
Because the answer must be a positive integer, it is just equal to the modulus of the product. Define $z_n = \left(\textrm{cis }\frac{2n\pi}{7}\right)^3 + \textrm{cis }\frac{2n\pi}{7} + 1$
Then, our product is equal to
\[|z_0||z_1||z_2||z_3||z_4||z_5||z_6|.\]
$z_0 = 0$ , and we may observe that $z_x$ and $z_{7-x}$ are conjugates for any $x$ , meaning that their magnitudes are the same. Thus, our product is
\[3|z_1|^2|z_2|^2|z_3|^2\] \[= 3\left((\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7} + 1)^2 + (\sin \frac{6\pi}{7} + \sin \frac{2\pi}{7})^2\right) \left((\cos \frac{12\pi}{7} + \cos \frac{4\pi}{7} + 1)^2 + (\sin \frac{12\pi}{7} + \sin \frac{4\pi}{7})^2\right) \left((\cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} + 1)^2 + (\sin \frac{4\pi}{7} + \sin \frac{6\pi}{7})^2\right)\]
Let us simplify the first term. Expanding, we obtain
\[\cos^2 \frac{6\pi}{7} + \cos^2 \frac{2\pi}{7} + 1 + 2\cos \frac{6\pi}{7}\cos \frac{2\pi}{7} + 2\cos \frac{6\pi}{7} + 2\cos \frac{2\pi}{7} + \sin^2 \frac{6\pi}{7} + \sin^2 \frac{2\pi}{7} + 2\sin \frac{6\pi}{7}\sin \frac{2\pi}{7}.\]
Rearranging and cancelling, we obtain
\[3 + 2\cos \frac{6\pi}{7} + 2\cos \frac{2\pi}{7} + 2\cos \frac{6\pi}{7}\cos \frac{2\pi}{7} + 2\sin \frac{6\pi}{7}\sin \frac{2\pi}{7}.\]
By the cosine subtraction formula, we have $2\cos \frac{6\pi}{7}\cos \frac{2\pi}{7} + 2\sin \frac{6\pi}{7}\sin \frac{2\pi}{7} = \cos \frac{6\pi - 2\pi}{7} = \cos \frac{4\pi}{7}$
Thus, the first term is equivalent to
\[3 + 2(\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}).\]
Similarly, the second and third terms are, respectively,
\[3 + 2(\cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} + \cos \frac{12\pi}{7}),\textrm{ and}\] \[3 + 2(\cos \frac{6\pi}{7} + \cos \frac{12\pi}{7} + \cos \frac{4\pi}{7}).\]
Next, we have $\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}$ . This is because
\[\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = \frac{1}{2}(\textrm{cis }\frac{2\pi}{7} + \textrm{cis }\frac{4\pi}{7} + \textrm{cis }\frac{6\pi}{7} + \textrm{cis }\frac{8\pi}{7} + \textrm{cis }\frac{10\pi}{7} + \textrm{cis }\frac{12\pi}{7})\]
\[= \frac{1}{2}(-1)\] \[= -\frac{1}{2}.\]
Therefore, the first term is simply $2$ . We have $\cos x = \cos 2\pi - x$ , so therefore the second and third terms can both also be simplified to $3 + 2(\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}) = 2$ . Thus, our answer is simply
\[3 \cdot 2 \cdot 2 \cdot 2\] \[= \boxed{024}.\]

Solution:
We write out the product in terms of $\omega$ \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=3(\omega^3+\omega+1)(\omega^6+\omega^2+1)(\omega^9+\omega^3+1)(\omega^{12}+\omega^4+1)(\omega^{15}+\omega^5+1)(\omega^{18}+\omega^6+1).\]
Grouping the terms in the following way exploits the fact that $\omega^{7k}=1$ for an integer $k$ , when multiplying out two adjacent products from left to right:
\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=(\omega^3+\omega+1)(\omega^{18}+\omega^6+1)(\omega^6+\omega^2+1)(\omega^{15}+\omega^5+1)(\omega^9+\omega^3+1)(\omega^{12}+\omega^4+1).\]
When multiplying two numbers with like bases, we add the exponents. We can now rewrite the exponents of each product (two at a time, where $1$ is treated as the identity) as a series of arrays:
\[\textbf{(A)}\begin{bmatrix} 3&1 &0 \\ 18&6&0\\ \end{bmatrix}\]
\[\textbf{(B)}\begin{bmatrix} 6&2 &0 \\ 15&5&0\\ \end{bmatrix}\]
\[\textbf{(C)}\begin{bmatrix} 9&3 &0 \\ 12&4&0\\ \end{bmatrix}.\]
Note that $\omega=e^{\frac{2\pi i}{7}}$ . When raising $\omega$ to a power, the numerator of the fraction is $2$ times whatever power $\omega$ is raised to, multiplied by $\pi i$ . Since the period of $\omega$ is $2\pi,$ we multiply each array by $2$ then reduce each entry $\mod{14},$ as each entry in an array represents an exponent which $\omega$ is raised to.
\[\textbf{(A)}\begin{bmatrix} 6&2 &0 \\ 8&12&0\\ \end{bmatrix}\]
\[\textbf{(B)}\begin{bmatrix} 12&4 &0 \\ 2&10&0\\ \end{bmatrix}\]
\[\textbf{(C)}\begin{bmatrix} 4&6 &0 \\ 10&8&0\\ \end{bmatrix}.\]
To obtain the correct exponents, we seperately add each element of the lower row to one element of the top row.
Therefore (after reducing $\mod 14$ again), we get the following sets:
\[\textbf{(A)}\ \{0, 4, 6, 10, 0, 2, 8, 12, 0\}\] \[\textbf{(B)}\ \{0, 8, 12, 6, 0, 4, 2, 10, 0\}\] \[\textbf{(C)}\ \{0, 12, 4, 2, 0, 8, 10, 8, 0\}.\]
Raising $\omega$ to the power of each element in every set then multiplying over $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}$ yields
\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=\left(\sum_{a\in \textbf{(A)}} \omega^a\right)\left(\sum_{b\in \textbf{(B)}} \omega^b\right)\left(\sum_{c\in \textbf{(C)}} \omega^c\right)\]
\[=\left(\sum_{a\in \textbf{(A)}} \omega^a\right)^3\]
\[=\left(\omega^0+\omega^4+\omega^6+\omega^{10}+\omega^0+\omega^2+\omega^8+\omega^{12}+\omega^0\right)^3\]
\[=\left(3+\omega^2+\omega^4+\omega^6+\omega^8+\omega^{10}+\omega^{12}\right)^3,\] as these sets are all identical.
Summing as a geometric series,
\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=\left(3+\frac{\omega^2(\omega^{12}-1)}{\omega^2-1}\right)^3\]
\[=\left(3+\frac{\omega^{14}-\omega^2}{\omega^2-1}\right)^3\]
\[=\left(3+\frac{1-\omega^2}{\omega^2-1}\right)^3\]
\[=(3-1)^3=8.\]
Therefore,
\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=8,\] and \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=3\cdot8=\boxed{024}.\]

Solution:
The product can be factored into $-(r-1)(s-1)(t-1)(r-w)(s-w)(t-w)(r-w^2)(s-w^2)(t-w^2)....(r-w^6)(s-w^6)(t-w^6)$
where $r,s,t$ are the roots of the polynomial $x^3+x+1=0$
This is then $-(r^7-1)(s^7-1)(t^7-1)$ because $(r^7-1)$ and $(r-1)(r-w)(r-w^2)...(r-w^6)$ share the same roots.
To find $-(r^7-1)(s^7-1)(t^7-1)$
Notice that $(r^7-1)=(r-1)(r^6+r^5+r^4+r^3+r^2+r+1)$ . Since r satisfies $x^3+x+1=0$ $r^6+r^4+r^3=0$
Substituting, you are left with $r^5+r^2+r+1$ . This is $r^2(r^3+1)+r+1$ , and after repeatedly substituting $r^3+x+1=0$ you are left with $-2r^3$
So now the problem is reduced to finding $-(r-1)(s-1)(t-1)(-2r^3)(-2s^3)(-2t^3)=8(rst)^3(r-1)(s-1)(t-1)$ , and vietas gives you the result of $\boxed{24}$ -resources


Answer: 321.0

Let $\omega\neq 1$ be a 13th root of unity. Find the remainder when \[\prod_{k=0}^{12}(2-2\omega^k+\omega^{2k})\] is divided by 1000.

Solution:
\[\prod_{k=0}^{12} \left(2- 2\omega^k + \omega^{2k}\right) = \prod_{k=0}^{12} \left((1 - \omega^k)^2 + 1\right) = \prod_{k=0}^{12} \left((1 + i) - \omega^k)((1 - i) - \omega^k\right)\]
Now, we consider the polynomial $x^{13} - 1$ whose roots are the 13th roots of unity. Taking our rewritten product from $0$ to $12$ , we see that both instances of $\omega^k$ cycle through each of the 13th roots. Then, our answer is:
\[((1 + i)^{13} - 1)(1 - i)^{13} - 1)\]
\[= (-64(1 + i) - 1)(-64(1 - i) - 1)\]
\[= (65 + 64i)(65 - 64i)\]
\[= 65^2 + 64^2\]
\[= 8\boxed{321}\]

Solution:
To find $\prod_{k=0}^{12} (2 - 2w^k + w^{2k})$ , where $w\neq1$ and $w^{13}=1$ , rewrite this is as
$(r-w)(s-w)(r-w^2)(s-w^2)...(r-w^{12})(s-w^{12})$ where $r$ and $s$ are the roots of the quadratic $x^2-2x+2=0$
Grouping the $r$ 's and $s$ 's results in $\frac{r^{13}-1}{r-1} \cdot\frac{s^{13}-1}{s-1}$
the denomiator $(r-1)(s-1)=1$ by vietas.
the numerator $(rs)^{13} - (r^{13} + s^{13}) + 1 = 2^{13} - (-128) + 1= 8321$ by newtons sums
so the answer is $\boxed{321}$

Solution:
Denote $r_j = e^{\frac{i 2 \pi j}{13}}$ for $j \in \left\{ 0, 1, \cdots , 12 \right\}$
Thus, for $\omega \neq 1$ $\left( \omega^0, \omega^1, \cdots, \omega^{12} \right)$ is a permutation of $\left( r_0, r_1, \cdots, r_{12} \right)$
We have
\begin{align*}\
\Pi_{k = 0}^{12} \left( 2 - 2 \omega^k + \omega^{2k} \right)
& = \Pi_{k=0}^{12} \left( 1 + i - \omega^k \right)
\left( 1 - i - \omega^k \right) \\
& = \Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - \omega^k \right)
\left( \sqrt{2} e^{-i \frac{\pi}{4}} - \omega^k \right) \\
& = \Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - r_k \right)
\left( \sqrt{2} e^{-i \frac{\pi}{4}} - r_k \right) \\
& = \left(
\Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - r_k \right)
\right)
\left(
\Pi_{k=0}^{12} \left( \sqrt{2} e^{-i \frac{\pi}{4}} - r_k \right)
\right) . \hspace{1cm} (1)
\end{align*}
The third equality follows from the above permutation property.
Note that $r_0, r_1, \cdots , r_{12}$ are all zeros of the polynomial $z^{13} - 1$ .
Thus, \[ z^{13} - 1 = \Pi_{k=0}^{12} \left( z - r_k \right) . \]
Plugging this into Equation (1), we get
\begin{align*}
(1)
& = \left( \left( \sqrt{2} e^{i \frac{\pi}{4}} \right)^{13} - 1 \right)
\left( \left( \sqrt{2} e^{-i \frac{\pi}{4}} \right)^{13} - 1 \right) \\
& = \left( - 2^{13/2} e^{i \frac{\pi}{4}} - 1 \right)
\left( - 2^{13/2} e^{-i \frac{\pi}{4}} - 1 \right) \\
& = 2^{13} + 1 + 2^{13/2} \cdot 2 \cos \frac{\pi}{4} \\
& = 2^{13} + 1 + 2^7 \\
& = 8321 .
\end{align*}
Therefore, the answer is $\boxed{321}$


Answer: 65.0

Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
 Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$ , respectively. There is a homothety at $A$ sending $\omega$ to $\omega_1$ that sends $B$ to $P$ and $O$ to $O_1$ , so $\overline{OO_2}\parallel\overline{O_1P}$ . Similarly, $\overline{OO_1}\parallel\overline{O_2P}$ , so $OO_1PO_2$ is a parallelogram. Moreover, \[\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,\] whence $OO_1O_2Q$ is cyclic. However, \[OO_1=O_2P=O_2Q,\] so $OO_1O_2Q$ is an isosceles trapezoid. Since $\overline{O_1O_2}\perp\overline{XY}$ $\overline{OQ}\perp\overline{XY}$ , so $Q$ is the midpoint of $\overline{XY}$
By Power of a Point, $PX\cdot PY=PA\cdot PB=15$ . Since $PX+PY=XY=11$ and $XQ=11/2$ \[XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,\] and the requested sum is $61+4=\boxed{065}$

Solution: One may solve for $PX$ first using PoAP, $PX = \frac{11}{2} - \frac{\sqrt{61}}{2}$ . Then, notice that $PQ^2$ is rational but $PX^2$ is not, also $PX = \frac{XY}{2} - \frac{\sqrt{61}}{2}$ . The most likely explanation for this is that $Q$ is the midpoint of $XY$ , so that $XQ = \frac{11}{2}$ and $PQ=\frac{\sqrt{61}}{2}$ . Then our answer is $m+n=61+4=\boxed{065}$ . One can rigorously prove this using the methods above

Solution: Let the tangents to $\omega$ at $A$ and $B$ intersect at $R$ . Then, since $RA^2=RB^2$ $R$ lies on the radical axis of $\omega_1$ and $\omega_2$ , which is $\overline{PQ}$ . It follows that \[-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).\] Let $Q'$ denote the midpoint of $\overline{XY}$ . By the Midpoint of Harmonic Bundles Lemma(EGMO 9.17), \[RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,\] whence $Q=Q'$ . Like above, $XP=\tfrac{11-\sqrt{61}}2$ . Since $XQ=\tfrac{11}2$ , we establish that $PQ=\tfrac{\sqrt{61}}2$ , from which $PQ^2=\tfrac{61}4$ , and the requested sum is $61+4=\boxed{065}$

Solution:
Firstly we need to notice that $Q$ is the middle point of $XY$ . Assume the center of circle $w, w_1, w_2$ are $O, O_1, O_2$ , respectively. Then $A, O_1, O$ are collinear and $O, O_2, B$ are collinear. Link $O_1P, O_2P, O_1Q, O_2Q$ . Notice that, $\angle B=\angle A=\angle APO_1=\angle BPO_2$ . As a result, $PO_1\parallel O_2O$ and $OO_1\parallel O_2P$ . So we have parallelogram $PO_2OO_1$ . So $\angle O_2PO_1=\angle O$ Notice that, $O_1O_2\bot PQ$ and $O_1O_2$ divides $PQ$ into two equal length pieces, So we have $\angle O_2QO_1=\angle O_2PO_1=\angle O$ . As a result, $O_2, Q, O, O_1,$ lie on one circle. So $\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P$ . Notice that since $\angle O_1PQ+\angle O_2O_1P=90^{\circ}$ , we have $\angle OQP=\angle OQO_1 + \angle O_1QP = \angle O_2O_1P + O_1PQ=90^{\circ}$ . As a result, $OQ\bot PQ$ . So $Q$ is the middle point of $XY$
Back to our problem. Assume $XP=x$ $PY=y$ and $x<y$ . Then we have $AP\cdot PB=XP\cdot PY$ , that is, $xy=15$ . Also, $XP+PY=x+y=XY=11$ . Solve these above, we have $x=\frac{11-\sqrt{61}}{2}=XP$ . As a result, we have $PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}$ . So, we have $PQ^2=\frac{61}{4}$ . As a result, our answer is $m+n=61+4=\boxed{065}$

Solution:
Note that the tangents to the circles at $A$ and $B$ intersect at a point $Z$ on $XY$ by radical axis theorem. Since $\angle ZAB = \angle ZQA$ and $\angle ZBA = \angle ZQB$ , we have \[\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},\] so $ZAQB$ is cyclic.
But if $O$ is the center of $\omega$ , clearly $ZAOB$ is cyclic with diameter $ZO$ , so $\angle ZQO = 90^{\circ}$ implies that $Q$ is the midpoint of $XY$ . Then, by power of point $P$ \[PY \cdot PX = PA \cdot PB = 15,\] whereas it is given that $PY+PX = 11$ . Thus \[PY, PX \in \left\{\tfrac 12 (11 \pm \sqrt{61})\right\}\] so $PQ = \frac{\sqrt{61}}{2}$ , i.e. $PQ^2 = \frac{61}{4}$ and the answer is $61+4 = \boxed{065}$

Solution:
Connect $AQ,QB$ , since $\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}$ , so $\angle{AQP}=\frac{\angle{AO_1P}}{2}=\angle{BQP}=\frac{\angle{BO_2P}}{2}, \angle{AQB}=\angle{AOB}$ then, so $A,O,Q,B$ are concyclic
We let $\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}=2\alpha$ , it is clear that $\angle{BQP}=\alpha, \angle{O_1AP}=90^{\circ}-\alpha$ , which leads to the conclusion $OQ\bot XY$ which tells $Q$ is the midpoint of $XY$
Then it is clear, $XP\cdot PY=15, XP=\frac{11-\sqrt{61}}{2}, PQ=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}$ , the answer is $\boxed{065}$

Solution: $PX \cdot PY=AP \cdot PB=5 \cdot 3=15$ by power of a point. Also, $PX+PY=XY=11$ , so $PX$ and $PY$ are solutions to the quadratic $x^2-11x+15=0$ so $PX$ and $PY$ is $\frac{11\pm\sqrt{61}}{2}$ in some order. Now, because we want $PQ^2$ and it is known to be rational, we can guess that $PQ$ is irrational or the problem would simply ask for $PQ$ . We can also figure out that since $PQ^2$ is rational, $PQ$ is $\sqrt{\text{[something]}}$ $PQ=QX-PX$ , and chances are low that $QX$ is some number with a square root plus or minus $\frac{\sqrt{61}}{2}$ to cancel out the $\frac{\sqrt{61}}{2}$ in $PX$ , so one can see that $PQ^2$ is most likely to be $\left(\frac{\sqrt{61}}{2}\right)^2=\frac{61}{4}$ , and our answer is $61+4=\boxed{065}$

Solution: We will show that $Q$ is the midpoint of $XY.$ To do this, let $Q^{\prime}$ be the altitude from $O$ to $XY$ or, equivalently, to $PQ.$ Notice that $O_{1}00_{2}P$ is a parallelogram. Thus, the height from $O$ to $O_{1}O_{2}$ is equal to the height from $P$ to $O_{1}O_{2}.$ Say that the line through $P$ perpendicular to $O_{1}O_{2}$ intersects $O_{1}O_{2}$ at $H.$ Then, $PQ$ is perpendicular to $O_{1}O_{2},$ so $H$ is on $PQ.$ Now, we have that the altitude from $O$ to $O_{1}O_{2}$ is equal to the altitude from $Q^{\prime}$ to $O_{1}O_{2}$ (since $OQ^{\prime} \parallel O_{1}O_{2}$ ). However, the altitude from $Q^{\prime}$ to $O_{1}O_{2}$ is just $Q^{\prime}H.$ Also, the altitude from $P$ to $O_{1}O_{2}$ is $PH$ , so $PH = Q^{\prime}H.$ Thus, $O_{1}O_{2}$ bisects $PQ^{\prime}.$ However, this is true for $Q,$ too, so $Q = Q^{\prime},$ and we are done. Now, by PoP, we have \[AP \cdot BP = XP \cdot YP = 15.\] Also, we have $XY = XP+YP = 11,$ so $XP = \frac{11 \pm \sqrt{61}}{2}$ . Notice that $XQ = \frac{XY}{2} = \frac{11}{2},$ so $PQ = \frac{\sqrt{61}}{2},$ giving us our answer of $\boxed{065}.$

Solution: Like Solution 7, let $Q'$ be the altitude from $O$ to $XY$ . And, let $M$ be the intersection of $O_1O_2$ and $PQ$ . Construct $P'$ on line $AO$ such that $PP' \parallel O_2O_1$ . First, because of isosceles triangles $OAB$ $O_1AP$ , and $O_2BP$ , we have $\angle{OAP} = \angle{OBA} = \angle{APO_1} = \angle{BPO_2}$ , which means $OO_1PO_2$ is a parallelogram. So, $O_2P = OO_1$ . It is also clear that $PP'O_1O_2$ is a parallelogram by virtue of our definition. Thus, $O_2P = O_1P' = OO_1$ . Since $OQ' \parallel O_1O_2 \parallel P'P$ (because of the right angles), $\frac{Q'M}{MP} = \frac{OO_1}{O_1P'} = 1 \implies Q'M = MP$ . And, because $QM = MP$ $Q = Q'$ . From Power of a Point on $P$ , we have $XP(11-XP) = 15$ , giving us $XP = \frac{11 - \sqrt{61}}{2}$ . Since $OQ$ is perpendicular to $XY$ $Q$ is the midpoint of $XY$ , so $XQ = \frac{11}{2}$ . Thus, $PQ = \frac{11}{2} - \frac{11 - \sqrt{61}}{2} = \frac{\sqrt{61}}{2} \implies {PQ}^2 = \frac{61}{4}$ . Therefore, our answer is $\boxed{65}$


Answer: 100.0

Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\  44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$

Solution: Let $O$ be the center of the circle, and $X$ be the midpoint of $\overline{CD}$ . Let $CX=a$ and $EX=b$ . This implies that $DE = a - b$ . Since $CE = CX + EX = a + b$ , we now want to find $(a+b)^2+(a-b)^2=2(a^2+b^2)$ . Since $\angle CXO$ is a right angle, by Pythagorean theorem $a^2 + b^2 = CX^2 + OX^2 = (5\sqrt{2})^2=50$ . Thus, our answer is $2\times50=\boxed{100}$

Solution:
Let $O$ be the center of the circle, and $X$ be the midpoint of $CD$ . Draw triangle $OCD$ , and median $OX$ . Because $OC = OD$ $OCD$ is isosceles, so $OX$ is also an altitude of $OCD$ $OE = 5\sqrt2 - 2\sqrt5$ , and because angle $OEC$ is $45$ degrees and triangle $OXE$ is right, $OX = EX = \frac{5\sqrt2 - 2\sqrt5}{\sqrt2} = 5 - \sqrt{10}$ . Because triangle $OXC$ is right, $CX = \sqrt{(5\sqrt2)^2 - (5 - \sqrt{10})^2} = \sqrt{15 + 10\sqrt{10}}$ . Thus, $CD = 2\sqrt{15 + 10\sqrt{10}}$
We are looking for $CE^2$ $DE^2$ which is also $(CE + DE)^2 - 2 \cdot CE \cdot DE$
Because $CE + DE = CD = 2\sqrt{15 + 10\sqrt{10}}$ $(CE + DE)^2 = CD^2=4(15 + 10\sqrt{10}) = 60 + 40\sqrt{10}$
By Power of a Point, $CE \cdot DE = AE \cdot BE = 2\sqrt5\cdot(10\sqrt2 - 2\sqrt5) = 20\sqrt{10} - 20$ , so $2 \cdot CE \cdot DE = 40\sqrt{10} - 40$
Finally, $CE^2 + DE^2 = (CE+ED)^2-2\cdot CE \cdot DE=(60 + 40\sqrt{10}) - (40\sqrt{10} - 40) = \boxed{100}$

Solution:
Let $O$ be the center of the circle. Notice how $OC = OD = r$ , where $r$ is the radius of the circle. By applying the law of cosines on triangle $OCE$ \[r^2=CE^2+OE^2-2(CE)(OE)\cos{45}=CE^2+OE^2-(CE)(OE)\sqrt{2}.\]
Similarly, by applying the law of cosines on triangle $ODE$ \[r^2=DE^2+OE^2-2(DE)(OE)\cos{135}=DE^2+OE^2+(DE)(OE)\sqrt{2}.\]
By subtracting these two equations, we get \[CE^2-DE^2-(CE)(OE)\sqrt{2}-(DE)(OE)\sqrt{2}=0.\] We can rearrange it to get \[CE^2-DE^2=(CE)(OE)\sqrt{2}+(DE)(OE)\sqrt{2}=(CE+DE)(OE\sqrt{2}).\]
Because both $CE$ and $DE$ are both positive, we can safely divide both sides by $(CE+DE)$ to obtain $CE-DE=OE\sqrt{2}$ . Because $OE = OB - BE = 5\sqrt{2} - 2\sqrt{5}$ \[(CE-DE)^2 = CE^2+DE^2 - 2(CE)(DE) = (OE\sqrt{2})^2 =2(5\sqrt{2} - 2\sqrt{5})^2 = 140 - 40\sqrt{10}.\]
Through power of a point, we can find out that $(CE)(DE)=20\sqrt{10} - 20$ , so \[CE^2+DE^2 = (CE-DE)^2+ 2(CE)(DE)= (140 - 40\sqrt{10}) + 2(20\sqrt{10} - 20) = \boxed{100}.\]

Solution:
 Let $O$ be the center of the circle. By reflecting $D$ across the line $AB$ to produce $D'$ , we have that $\angle BED'=45$ . Since $\angle AEC=45$ $\angle CED'=90$ . Since $DE=ED'$ , by the Pythagorean Theorem, our desired solution is just $CD'^2$ .
Looking next to circle arcs, we know that $\angle AEC=\frac{\overarc{AC}+\overarc{BD}}{2}=45$ , so $\overarc{AC}+\overarc{BD}=90$ . Since $\overarc{BD'}=\overarc{BD}$ , and $\overarc{AC}+\overarc{BD'}+\overarc{CD'}=180$ $\overarc{CD'}=90$ . Thus, $\angle COD'=90$ .
Since $OC=OD'=5\sqrt{2}$ , by the Pythagorean Theorem, the desired $CD'^2= \boxed{100}$

Solution:
Basically, by PoP, you have that \[CE \times DE = (10\sqrt{2}-2\sqrt{5})(2\sqrt{5}) = 20\sqrt{10} - 20.\] Therefore, as $CE^2 + DE^2 = (CD)^2 - 2(CE \times DE),$ basically, once you find $CD^2,$ the problem is done. Now, this is an IMPORTANT concept: If you have a circle which you know the radius of and you want to find the length of a chord of that circle, drop an altitude from the center of the circle to the chord to find distance between the center of the circle and the chord.
In this case, let $M$ be the midpoint of chord $CD.$ Notice that now we can use our $45^\circ{}$ angle, since $OME$ is a $45^\circ{}-45^\circ{}-90^\circ{}$ triangle so that $ME = x$ and $OE = x\sqrt{2}.$ However, we have that $OE = 5\sqrt{2}-2\sqrt{5},$ so that $x = 5 - \sqrt{10}.$ Now, notice that $x^2 = 35 - 10\sqrt{10},$ so that \[CM^2 = 50 - x^2 = 50 - (35 - 10\sqrt{10}) = 15 + 10 \sqrt{10}\] and \[CD^2 = 60 + 4\sqrt{10}.\] Therefore, \[CE^2 + DE^2 = CD^2 - 2(CE \times DE) = (60+4\sqrt{10}) - 2(2\sqrt{10} - 20) = (60+4\sqrt{10}) + (40-4\sqrt{10}) = \boxed{100}.\]

Solution:
 For ease of notation, let $x = DE$ and $y=EC$ .
Extend $\overline{CD}$ to point $P$ until $\overline{BP}$ is perpendicular to $AB$ . It's given that $\angle AEC = 45^{\circ}$ , so, by vertical angles, we have $\angle BEP = \angle EPB = 45^{\circ}$
Since $PEB$ is a $45-45-90$ right triangle, we have $BE = BP = 2\sqrt{5}$ and $PE=2\sqrt{10}$ . Hence, $PD = 2\sqrt{10}-x.$
By Power of a Point, we have
\[PB^2 = PD\cdot PC\] \[20 = \left(2\sqrt{10}-x\right)\left(y + 2\sqrt{10}\right).\]
Isolating the variables after expanding gives $x-y = 2\sqrt{10}-10.$
Using Power of a Point again, we have
\[DE\cdot EC = BE\cdot EA\] \[xy = 2\sqrt{5}\left(10\sqrt{2}-2\sqrt{5} \right)\] \[xy = 20\sqrt{10}-20.\] To get $x^2 + y^2$ , we can perform the operation $(x-y)^2 + 2xy$ . Plugging these values in,
\[(x-y)^2 + 2xy = \left(2\sqrt{10}-10\right)^2 + 2\left(20\sqrt{10}-20\right)\] \[= 40 + 100 - 40\sqrt{10} + 40\sqrt{10} - 40 = \boxed{100}\]

Solution:
Perhaps not reliable in general, but very useful as a last resort.
The choice of the radius $5\sqrt2$ is strange, and is probably motivated by a nice answer in the end, so we only consider integer options. Notice that a 5 also appears in the condition $BE=2\sqrt5$ , therefore it will likely be present in the answer as well; the only integer containing a factor of 5 amongst the choices is 100, thus the answer is $\boxed{100}$


Answer: 432.0

Let $\overline{AB}$ be a diameter of circle $\omega$ . Extend $\overline{AB}$ through $A$ to $C$ . Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$ . Point $P$ is the foot of the perpendicular from $A$ to line $CT$ . Suppose $\overline{AB} = 18$ , and let $m$ denote the maximum possible length of segment $BP$ . Find $m^{2}$

Solution: Let $x = OC$ . Since $OT, AP \perp TC$ , it follows easily that $\triangle APC \sim \triangle OTC$ . Thus $\frac{AP}{OT} = \frac{CA}{CO} \Longrightarrow AP = \frac{9(x-9)}{x}$ . By the Law of Cosines on $\triangle BAP$ \begin{align*}BP^2 = AB^2 + AP^2 - 2 \cdot AB \cdot AP \cdot \cos \angle BAP \end{align*} where $\cos \angle BAP = \cos (180 - \angle TOA) = - \frac{OT}{OC} = - \frac{9}{x}$ , so: \begin{align*}BP^2 &= 18^2 + \frac{9^2(x-9)^2}{x^2} + 2(18) \cdot \frac{9(x-9)}{x} \cdot \frac 9x = 405 + 729\left(\frac{2x - 27}{x^2}\right)\end{align*} Let $k = \frac{2x-27}{x^2} \Longrightarrow kx^2 - 2x + 27 = 0$ ; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(k)(27) \ge 0 \Longleftrightarrow k \le \frac{1}{27}$ . Thus, \[BP^2 \le 405 + 729 \cdot \frac{1}{27} = \boxed{432}\] Equality holds when $x = 27$

Solution:
From the diagram, we see that $BQ = OT + BO \sin\theta = 9 + 9\sin\theta = 9(1 + \sin\theta)$ , and that $QP = BA\cos\theta = 18\cos\theta$
\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \sin\theta)^2 + 18^2\cos^2\theta\\ &= 9^2[1 + 2\sin\theta + \sin^2\theta + 4(1 - \sin^2\theta)]\\ BP^2 &= 9^2[5 + 2\sin\theta - 3\sin^2\theta]\end{align*}
This is a quadratic equation , maximized when $\sin\theta = \frac { - 2}{ - 6} = \frac {1}{3}$ . Thus, $m^2 = 9^2[5 + \frac {2}{3} - \frac {1}{3}] = \boxed{432}$

Solution:
(Diagram credit goes to Solution 2)
We let $AC=x$ . From similar triangles, we have that $PC=\frac{x\sqrt{x^2+18x}}{x+9}$ (Use Pythagorean on $\triangle\omega TC$ and then using $\triangle\omega CT\sim\triangle ACP$ ). Similarly, $TP=QT=\frac{9\sqrt{x^2+18x}}{x+9}$ . Using the Pythagorean Theorem again and $\triangle CAP\sim\triangle CBQ$ $BQ=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2}$ . Using the Pythagorean Theorem $\bold{again}$ $BP=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2+(\frac{18\sqrt{x^2+18x}}{x+9})^2}$ . After a large bashful simplification, $BP=\sqrt{405+\frac{1458x-6561}{x^2+18x+81}}$ . The fraction is equivalent to $729\frac{2x-9}{(x+9)^2}$ . Taking the derivative of the fraction and solving for x, we get that $x=18$ . Plugging $x=18$ back into the expression for $BP$ yields $\sqrt{432}$ , so the answer is $(\sqrt{432})^2=\boxed{432}$

Solution:
Let $h$ be the distance from $A$ to $CT$ . Observe that $h$ takes any value from $0$ to $2r$ , where $r$ is the radius of the circle.
Let $Q$ be the foot of the altitude from $B$ to $CT$ . It is clear that $T$ is the midpoint of $PQ$ , and so the length $OT$ is the average of $AP$ and $BQ$ . It follows thus that $BQ = 2r - h$
We compute $PT = \sqrt{r^2 - (r - h)^2} = \sqrt{h(2r - h)},$ and so $BP^2 = PQ^2 + BQ^2 = 4PT^2 + BQ^2 = 4h(2r - h) + (2r-h)^2 = (2r-h)(2r + 3h)$ .
This is $\frac{1}{3}(6r - 3h)(2r + 3h) \le \frac{1}{3} \cdot \left( \frac{8r}{2} \right)^2$ . Equality is attained, so thus we extract the answer of $\frac{16 \cdot 9^2}{3} = 27 \cdot 16 = \boxed{432}.$


Answer: 997.0

Let $\overline{CH}$ be an altitude of $\triangle ABC$ . Let $R\,$ and $S\,$ be the points where the circles inscribed in the triangles $ACH\,$ and $BCH^{}_{}$ are tangent to $\overline{CH}$ . If $AB = 1995\,$ $AC = 1994\,$ , and $BC = 1993\,$ , then $RS\,$ can be expressed as $m/n\,$ , where $m\,$ and $n\,$ are relatively prime integers. Find $m + n\,$

Solution:
From the Pythagorean Theorem $AH^2+CH^2=1994^2$ , and $(1995-AH)^2+CH^2=1993^2$
Subtracting those two equations yields $AH^2-(1995-AH)^2=3987$
After simplification, we see that $2*1995AH-1995^2=3987$ , or $AH=\frac{1995}{2}+\frac{3987}{2*1995}$
Note that $AH+BH=1995$
Therefore we have that $BH=\frac{1995}{2}-\frac{3987}{2*1995}$
Therefore $AH-BH=\frac{3987}{1995}$
Now note that $RS=|HR-HS|$ $RH=\frac{AH+CH-AC}{2}$ , and $HS=\frac{CH+BH-BC}{2}$
Therefore we have $RS=\left| \frac{AH+CH-AC-CH-BH+BC}{2} \right|=\frac{|AH-BH-1994+1993|}{2}$
Plugging in $AH-BH$ and simplifying, we have $RS=\frac{1992}{1995*2}=\frac{332}{665} \rightarrow 332+665=\boxed{997}$


Answer: 14.0

Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$ . Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with chords $\overline{AC}$ and $\overline{BC}$ . The largest possible value of $d$ can be written in the form $r-s\sqrt{t}$ , where $r, s$ and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$

Solution:
Let $V = \overline{NM} \cap \overline{AC}$ and $W = \overline{NM} \cap \overline{BC}$ . Further more let $\angle NMC = \alpha$ and $\angle MNC = 90^\circ - \alpha$ . Angle chasing reveals $\angle NBC = \angle NAC = \alpha$ and $\angle MBC = \angle MAC = 90^\circ - \alpha$ . Additionally $NB = \frac{4}{5}$ and $AN = AM$ by the Pythagorean Theorem.
By the Angle Bisector Formula, \[\frac{NV}{MV} = \frac{\sin (\alpha)}{\sin (90^\circ - \alpha)} = \tan (\alpha)\] \[\frac{MW}{NW} = \frac{3\sin (90^\circ - \alpha)}{4\sin (\alpha)} = \frac{3}{4} \cot (\alpha)\]
As $NV + MV =MW + NW = 1$ we compute $NW = \frac{1}{1+\frac{3}{4}\cot(\alpha)}$ and $MV = \frac{1}{1+\tan (\alpha)}$ , and finally $VW = NW + MV - 1 =  \frac{1}{1+\frac{3}{4}\cot(\alpha)} + \frac{1}{1+\tan (\alpha)} - 1$ . Taking the derivative of $VW$ with respect to $\alpha$ , we arrive at \[VW' = \frac{7\cos^2 (\alpha) - 4}{(\sin(\alpha) + \cos(\alpha))^2(4\sin(\alpha)+3\cos(\alpha))^2}\] Clearly the maximum occurs when $\alpha = \cos^{-1}\left(\frac{2}{\sqrt{7}}\right)$ . Plugging this back in, using the fact that $\tan(\cos^{-1}(x)) = \frac{\sqrt{1-x^2}}{x}$ and $\cot(\cos^{-1}(x)) = \frac{x}{\sqrt{1-x^2}}$ , we get
$VW = 7 - 4\sqrt{3}$ with $7 + 4 + 3 = \boxed{014}$

Solution:
Since $MA = \frac{\sqrt{2}}{2} \approx 0.707 > \frac{3}{5}$ , point $B$ lies between $M$ and $A$ on the semicircular arc. We will first compute the length of $\overline{AB}$ . By the law of cosines, $\cos \angle MOB = \frac{-(3/5)^2 + 2(1/2)^2}{2(1/2)^2} = \frac{7}{25}$ , so $\cos \angle AOB = \sin \angle MOB = \frac{24}{25}$ . Then $AB^2 = 2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right)^2 \cdot \frac{24}{25} = \frac{1}{50}$ , so $AB = \frac{1}{5\sqrt{2}}$
Let $P = AC \cap MN$ and $Q = BC \cap MN$ , and let $MQ = x$ $PQ = d$ $PN = y$ . Note that \[(M, P; Q, N) \stackrel{C}{=} (M, A; B, N),\] that is, \[\frac{QP}{QM} \div \frac{NP}{NM} = \frac{BA}{BM} \div \frac{NA}{NM}\] or \[\frac{d}{xy} = \frac{1/(5\sqrt{2})}{(3/5) \cdot (\sqrt{2}/2)} = \frac{1}{3}.\] Hence $d = \frac{1}{3}xy$ , and we also know $d+x+y=1$ . Now AM-GM gives \[\frac{x+y}{2} \ge \sqrt{xy} \implies \frac{1-d}{2} \ge \sqrt{3d}.\] This gives the quadratic inequality $d^2 - 14d + 1 \ge 0$ , which solves as \[d \in \left(-\infty, 7-4\sqrt3\right] \cup \left[7+4\sqrt3, \infty\right).\] But $d \le 1$ , so the greatest possible value of $d$ is $7-4\sqrt3$ . The answer is $7+4+3=\boxed{014}$

Solution:
Let $O$ be the center of the circle. Define $\angle{MOC}=t$ $\angle{BOA}=2a$ , and let $BC$ and $AC$ intersect $MN$ at points $X$ and $Y$ , respectively. We will express the length of $XY$ as a function of $t$ and maximize that function in the interval $[0, \pi]$
Let $C'$ be the foot of the perpendicular from $C$ to $MN$ . We compute $XY$ as follows.
(a) By the Extended Law of Sines in triangle $ABC$ , we have
\[CA\]
\[= \sin\angle{ABC}\]
\[= \sin\left(\frac{\widehat{AN} + \widehat{NC}}{2}\right)\]
\[= \sin\left(\frac{\frac{\pi}{2} + (\pi-t)}{2}\right)\]
\[= \sin\left(\frac{3\pi}{4} - \frac{t}{2}\right)\]
\[= \sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\]
(b) Note that $CC' = CO\sin(t) = \left(\frac{1}{2}\right)\sin(t)$ and $AO = \frac{1}{2}$ . Since $CC'Y$ and $AOY$ are similar right triangles, we have $CY/AY = CC'/AO = \sin(t)$ , and hence,
\[CY/CA\]
\[= \frac{CY}{CY + AY}\]
\[= \frac{\sin(t)}{1 + \sin(t)}\]
\[= \frac{\sin(t)}{\sin\left(\frac{\pi}{2}\right) + \sin(t)}\]
\[= \frac{\sin(t)}{2\sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)}\]
(c) We have $\angle{XCY} = \frac{\widehat{AB}}{2}=a$ and $\angle{CXY} = \frac{\widehat{MB}+\widehat{CN}}{2} = \frac{\left(\frac{\pi}{2} - 2a\right) + (\pi - t)}{2} = \frac{3\pi}{4} - a - \frac{t}{2}$ , and hence by the Law of Sines,
\[XY/CY\]
\[= \frac{\sin\angle{XCY}}{\sin\angle{CXY}}\]
\[= \frac{\sin(a)}{\sin\left(\frac{3\pi}{4} - a - \frac{t}{2}\right)}\]
\[= \frac{\sin(a)}{\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}\]
(d) Multiplying (a), (b), and (c), we have
\[XY\]
\[= CA * (CY/CA) * (XY/CY)\]
\[= \frac{\sin(t)\sin(a)}{2\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}\]
\[= \frac{\sin(t)\sin(a)}{\sin\left(\frac{\pi}{2} + a\right) + \sin(a + t)}\]
\[= \sin(a)\times\frac{\sin(t)}{\sin(t + a) + \cos(a)}\]
which is a function of $t$ (and the constant $a$ ). Differentiating this with respect to $t$ yields
\[\sin(a)\times\frac{\cos(t)(\sin(t + a) + \cos(a)) - \sin(t)\cos(t + a)}{(\sin(t + a) + \cos(a))^2}\]
and the numerator of this is
\[\sin(a) \times(\sin(t + a)\cos(t) - \cos(t + a)\sin(t) + \cos(a)\cos(t))\] \[= \sin(a) \times (\sin(a) + \cos(a)\cos(t))\]
which vanishes when $\sin(a) + \cos(a)\cos(t) = 0$ . Therefore, the length of $XY$ is maximized when $t=t'$ , where $t'$ is the value in $[0, \pi]$ that satisfies $\cos(t') = -\tan(a)$
Note that
\[\frac{1 - \tan(a)}{1 + \tan(a)} = \tan\left(\frac{\pi}{4} - a\right) = \tan((\widehat{MB})/2) = \tan\angle{MNB} = \frac{3}{4}\]
so $\tan(a) = \frac{1}{7}$ . We compute
\[\sin(a) = \frac{\sqrt{2}}{10}\]
\[\cos(a) = \frac{7\sqrt{2}}{10}\]
\[\cos(t') = -\tan(a) = -\frac{1}{7}\]
\[\sin(t') = \frac{4\sqrt{3}}{7}\]
\[\sin(t' + a)=\sin(t')\cos(a) + \cos(t')\sin(a) = \frac{28\sqrt{6} - \sqrt{2}}{70}\]
so the maximum length of $XY$ is $\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt{3}$ , and the answer is $7 + 4 + 3 = \boxed{014}$

Solution:
By Pythagoras in $\triangle BMN,$ we get $BN=\dfrac{4}{5}.$
Since cross ratios are preserved upon projecting, note that $(M,Y;X,N)\stackrel{C}{=}(M,B;A,N).$ By definition of a cross ratio, this becomes \[\dfrac{XM}{NY}:\dfrac{NM}{NY}=\dfrac{AM}{AB}:\dfrac{MN}{NB}.\] Let $MY=a,YX=b,XN=c$ such that $a+b+c=1.$ We know that $XM=a+b,XY=b,NM=1,NY=b+c,$ so the LHS becomes $\dfrac{(a+b)(b+c)}{b}.$
In the RHS, we are given every value except for $AB.$ However, Ptolemy's Theorem on $MBAN$ gives $AB\cdot MN+AN\cdot BM=AM\cdot BN\implies AB+\dfrac{3}{5\sqrt{2}}=\dfrac{4}{5\sqrt{2}}\implies AB=\dfrac{1}{5\sqrt{2}}.$ Substituting, we get $\dfrac{(a+b)(b+c)}{b}=4\implies b(a+b+c)+ac=4b, b=\dfrac{ac}{3}$ where we use $a+b+c=1.$
Again using $a+b+c=1,$ we have $a+b+c=1\implies a+\dfrac{ac}{3}+c=1\implies a=3\dfrac{1-c}{c+3}.$ Then $b=\dfrac{ac}{3}=\dfrac{c-c^2}{c+3}.$ Since this is a function in $c,$ we differentiate WRT $c$ to find its maximum. By quotient rule, it suffices to solve \[(-2c+1)(c+3)-(c-c^2)=0 \implies c^2+6c-3,c=-3+2\sqrt{3}.\] Substituting back yields $b=7-4\sqrt{3},$ so $7+4+3=\boxed{014}$ is the answer.


Answer: 540.0

Let $\tau(n)$ denote the number of positive integer divisors of $n$ . Find the sum of the six least positive integers $n$ that are solutions to $\tau (n) + \tau (n+1) = 7$

Solution:
In order to obtain a sum of $7$ , we must have:
Since both of these cases contain a number with an odd number of divisors, that number must be an even power of a prime. These can come in the form of a square-like $3^2$ with $3$ divisors, or a fourth power like $2^4$ with $5$ divisors. We then find the smallest such values by hand.
Having computed the working possibilities, we take the sum of the corresponding values of $n$ $8+9+16+25+121+361 = \boxed{540}$ . ~Kepy.

Solution:
Let the ordered pair $(a,b)$ represent the number of divisors of $n$ and $n+1$ respectively.
We see that to obtain a sum of $7$ , we can have $(2,5), (3,4), (4,3),$ and $(5,2)$
Case 1: When we have $(2,5)$ For $n$ to have 2 divisors, it must be a prime number.
For $n+1$ to have 5 divisors, it must be in the form $a^4$ .
If $n+1$ is in the form $a^4$ , then $n = a^4-1 = (a^2+1)(a-1)(a+1)$ . This means that $n$ , or $a^4-1$ has factors other than 1 and itself; $n$ is not prime.
No cases work in this case
Case 2: When we have $(4,3)$ For $n$ to have 4 divisors, it must be in the form $a^3$ or $ab$ , where $a$ and $b$ are distinct prime numbers .
For $n+1$ to have 3 divisors, it must be a square number.
Let $n+1 = A^2$ $A$ is a prime number). When $n = a^3, a^3+1 = A^2, (A-1)(A+1)=a^3$ .
We see that the only case when it works is when $a=2, A=3$ , so $n=8$ works.
Case 3: When we have $(5,2)$ For $n$ to have 5 divisors, it must be in the form $a^4$ , where $a$ is a prime number.
For $n+1$ to have 2 divisors, it must be a prime number.
Notice that $a$ and $a^4$ have the same parity (even/odd). Since every prime greater than 2 are odd, $n = a^4$ must be even. Since $a^4$ is even, $a$ must be even as well, and the only prime number that is even is 2. When $a=2, n=16$
Case 4: When we have $(3,4)$ For $n$ to have 3 divisors, it must be a square number.
For $n+1$ to have 4 divisors, it must be in the form $a^3$ or $ab$ , where $a$ and $b$ are distinct prime numbers.
Similar to Case 2, let $n = A^2$ $A$ is a prime number).
There are no cases that satisfy this equation.
We test squares of primes to find values of n that work.
Now we add up the values of $n$ to get the answer: $8+16+9+25+121+361 = \boxed{540}$ .
~toastybaker


Answer: 18.0

Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$

Solution:
We have $\angle BCE = \angle ECD = \angle ECA = \tfrac 13 \cdot 90^\circ = 30^\circ$ . Drop the altitude from $D$ to $CB$ and call the foot $F$
Let $CD = 8a$ . Using angle bisector theorem on $\triangle CDB$ , we get $CB = 15a$ . Now $CDF$ is a $30$ $60$ $90$ triangle, so $CF = 4a$ $FD = 4a\sqrt{3}$ , and $FB = 11a$ .  Finally, $\tan{B} = \tfrac{DF}{FB}=\tfrac{4\sqrt{3}a}{11a} = \tfrac{4\sqrt{3}}{11}$ . Our final answer is $4 + 3 + 11 = \boxed{018}$

Solution:
Without loss of generality, set $CB = 1$ . Then, by the Angle Bisector Theorem on triangle $DCB$ , we have $CD = \frac{8}{15}$ . We apply the Law of Cosines to triangle $DCB$ to get $1 + \frac{64}{225} - \frac{8}{15} = BD^{2}$ , which we can simplify to get $BD = \frac{13}{15}$
Now, we have $\cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}$ by another application of the Law of Cosines to triangle $DCB$ , so $\cos \angle B = \frac{11}{13}$ . In addition, $\sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}$ , so $\tan \angle B = \frac{4\sqrt{3}}{11}$
Our final answer is $4+3+11 = \boxed{018}$

Solution:
(This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig)
Find values for all angles in terms of $\angle B$ $\angle CEB = 150-B$ $\angle CED = 30+B$ $\angle CDE = 120-B$ $\angle CDA = 60+B$ , and $\angle A = 90-B$
Use the law of sines on $\triangle CED$ and $\triangle CEB$
In $\triangle CED$ $\frac{8}{\sin 30} = \frac{CE}{\sin (120-B)}$ . This simplifies to $16 = \frac{CE}{\sin (120-B)}$
In $\triangle CEB$ $\frac{15}{\sin 30} = \frac{CE}{\sin B}$ . This simplifies to $30 = \frac{CE}{\sin B}$
Solve for $CE$ and equate them so that you get $16\sin (120-B) = 30\sin B$
From this, $\frac{8}{15} = \frac{\sin B}{\sin (120-B)}$
Use a trig identity on the denominator on the right to obtain: $\frac{8}{15} = \frac{\sin B}{\sin 120 \cos B - \cos 120 \sin B}$
This simplifies to $\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cos B + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}$
This gives $8\sqrt{3}\cos B+8\sin B=30\sin B$ Dividing by $\cos B$ , we have ${8\sqrt{3}}= 22\tan B$
$\tan{B} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}$ . Our final answer is $4 + 3 + 11 = \boxed{018}$

Solution:
(This solution avoids advanced trigonometry)
Let $X$ be the foot of the perpendicular from $D$ to $\overline{BC}$ , and let $Y$ be the foot of the perpendicular from $E$ to $\overline{BC}$
Now let $EY=x$ . Clearly, triangles $EYB$ and $DXB$ are similar with $\frac{BE}{BD}=\frac{15}{15+8}=\frac{15}{23}=\frac{EY}{DX}$ , so $DX=\frac{23}{15}x$
Since triangles $CDX$ and $CEY$ are 30-60-90 right triangles, we can easily find other lengths in terms of $x$ . For example, we see that $CY=x\sqrt{3}$ and $CX=\frac{\frac{23}{15}x}{\sqrt{3}}=\frac{23\sqrt{3}}{45}x$ . Therefore $XY=CY-CX=x\sqrt{3}-\frac{23\sqrt{3}}{45}x=\frac{22\sqrt{3}}{45}x$
Again using the fact that triangles $EYB$ and $DXB$ are similar, we see that $\frac{BX}{BY}=\frac{XY+BY}{BY}=\frac{XY}{BY}+1=\frac{23}{15}$ , so $BY=\frac{15}{8}XY=\frac{15}{8}*\frac{22\sqrt{3}}{45}=\frac{11\sqrt{3}}{2}$
Thus $\tan \angle B = \frac{x}{\frac{11\sqrt{3}}{12}x}=\frac{4\sqrt{3}}{11}$ , and our answer is $4+3+11=\boxed{018}$

Solution:
(Another solution without trigonometry)
Extend $CD$ to point $F$ such that $\overline{AF} \parallel \overline{CB}$ . It is then clear that $\triangle AFD$ is similar to $\triangle BCD$
Let $AC=p$ $BC=q$ . Then $\tan \angle B = p/q$
With the Angle Bisector Theorem, we get that $CD=\frac{8}{15}q$ . From 30-60-90 $\triangle CAF$ , we get that $AF=\frac{1}{\sqrt{3}}p$ and $FD=FC-CD=\frac{2}{\sqrt{3}}p-\frac{8}{15}q$
From $\triangle AFD \sim \triangle BCD$ , we have that $\frac{FD}{CD}=\frac{FA}{CB}=\frac{\frac{2}{\sqrt{3}}p-\frac{8}{15}q}{\frac{8}{15}q}=\frac{\frac{1}{\sqrt{3}}p}{q}$ . Simplifying yields $\left(\frac{p}{q}\right)\left(\frac{2\sqrt{3}}{3}*\frac{15}{8}-\frac{\sqrt{3}}{3}\right)=1$ , and $\tan \angle B=\frac{p}{q}=\frac{4\sqrt{3}}{11}$ , so our answer is $4+3+11=\boxed{018}$

Solution:
Let $CB = 1$ , and let the feet of the altitudes from $D$ and $E$ to $\overline{CB}$ be $D'$ and $E'$ , respectively. Also, let $DE = 8k$ and $EB = 15k$ . We see that $BD' = 15k\cos B$ and $BE' = 23k\cos B$ by right triangles $\triangle{BDD'}$ and $\triangle{BEE'}$ . From this we have that $D'E' = 8k\cos B$ . With the same triangles we have $DD' = 23k\sin B$ and $EE' = 15k\sin B$ . From 30-60-90 triangles $\triangle{CDD'}$ and $\triangle{CEE'}$ , we see that $CD' = \frac{23k\sqrt{3}\sin B}{3}$ and $CE' = 15k\sqrt{3}\sin B$ , so $D'E' = \frac{22k\sqrt{3}\sin B}{3}$ . From our two values of $D'E'$ we get: $8k\cos B = \frac{22k\sqrt{3}\sin B}{3}$
$\frac{\sin B}{\cos B} = \frac{8k}{\frac{22k\sqrt{3}}{3}} = \tan B$
$\tan B = \frac{8}{\frac{22\sqrt{3}}{3}} = \frac{24}{22\sqrt{3}} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}$ Our answer is then $4+3+11 = \boxed{018}$


Answer: 10.0

Let $\triangle ABC$ be a scalene triangle. Point $P$ lies on $\overline{BC}$ so that $\overline{AP}$ bisects $\angle BAC.$ The line through $B$ perpendicular to $\overline{AP}$ intersects the line through $A$ parallel to $\overline{BC}$ at point $D.$ Suppose $BP=2$ and $PC=3.$ What is $AD?$
$\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$

Solution:
Suppose that $\overline{BD}$ intersects $\overline{AP}$ and $\overline{AC}$ at $X$ and $Y,$ respectively. By Angle-Side-Angle, we conclude that $\triangle ABX\cong\triangle AYX.$
Let $AB=AY=2x.$ By the Angle Bisector Theorem, we have $AC=3x,$ or $YC=x.$
By alternate interior angles, we get $\angle YAD=\angle YCB$ and $\angle YDA=\angle YBC.$ Note that $\triangle ADY \sim \triangle CBY$ by the Angle-Angle Similarity, with the ratio of similitude $\frac{AY}{CY}=2.$ It follows that $AD=2CB=2(BP+PC)=\boxed{10}.$

Solution:
Let the intersection of $AC$ and $BD$ be $M$ , and the intersection of $AP$ and $BD$ be $N$ . Draw a line from $M$ to $BC$ , and label the point of intersection $O$
By adding this extra line, we now have many pairs of similar triangles. We have $\triangle BPN \sim \triangle BOM$ , with a ratio of $2$ , so $BO = 4$ and $OC = 1$ . We also have $\triangle COM \sim \triangle CAP$ with ratio $3$ . Additionally, $\triangle BPN \sim \triangle ADN$ (with an unknown ratio). It is also true that $\triangle BAN \cong \triangle MAN$
Suppose the area of $\triangle COM$ is $x$ . Then, $[\triangle CAP] = 9x$ . Because $\triangle CAP$ and $\triangle BAP$ share the same height and have a base ratio of $3:2$ $[\triangle BAP] = 6x$ . Because $\triangle BOM$ and $\triangle COM$ share the same height and have a base ratio of $4:1$ $[\triangle BOM] = 4x$ $[\triangle BPN] = x$ , and thus $[OMNP] = 4x - x = 3x$ . Thus, $[\triangle MAN] = [\triangle BAN] = 5x$
Finally, we have $\frac{[\triangle BAN]}{[\triangle BPN]} = \frac{5x}{x} = 5$ , and because these triangles share the same height $\frac{AN}{PN} = 5$ . Notice that these side lengths are corresponding side lengths of the similar triangles $BPN$ and $ADN$ . This means that $AD = 5\cdot BP = \boxed{10}$

Solution:
Let point $B$ be the origin, with $C$ being on the positive $x$ -axis and $A$ being in the first quadrant.
By the Angle Bisector Theorem, $AB:AC = 2:3$ . Thus, assume that $AB = 4$ , and $AC = 6$
Let the perpendicular from $A$ to $BC$ be $AM$
Using Heron's formula, \[[ABC] = \sqrt{\frac{15}{2}\left(\frac{15}{2}-4\right)\left(\frac{15}{2}-5\right)\left(\frac{15}{2}-6\right)} = \frac{15}{4}\sqrt{7}.\]
Hence, \[AM = \frac{\frac{15}{4}\sqrt{7}}{\frac{5}{2}} = \frac{3}{2}\sqrt{7}.\]
Next, we have \[BM^2 + AM^2 = AB^2\] \[\therefore BM = \sqrt{16 - \frac{63}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \textrm{ and } PM = \frac{3}{2}.\]
The slope of line $AP$ is thus \[\frac{-\frac{3}{2}\sqrt{7}}{\frac{3}{2}} = -\sqrt{7}.\]
Therefore, since the slopes of perpendicular lines have a product of $-1$ , the slope of line $BD$ is $\frac{1}{\sqrt{7}}$ . This means that we can solve for the coordinates of $D$
\[y = \frac{3}{2}\sqrt{7}\] \[y = \frac{1}{\sqrt{7}}x\] \[\frac{1}{\sqrt{7}}x = \frac{3}{2}\sqrt{7}\] \[x = \frac{7 \cdot 3}{2} = \frac{21}{2}\] \[D = \left(\frac{21}{2}, \frac{3}{2}\sqrt{7}\right).\]
We also know that the coordinates of $A$ are $\left(\frac{1}{2}, \frac{3}{2}\sqrt{7}\right)$ , because $BM = \frac{1}{2}$ and $AM = \frac{3}{2}\sqrt{7}$
Since the $y$ -coordinates of $A$ and $D$ are the same, and their $x$ -coordinates differ by $10$ , the distance between them is $10$ . Our answer is $\boxed{10}.$

Solution: Since there is only one possible value of $AD$ , we assume $\angle{B}=90^{\circ}$ . By the angle bisector theorem, $\frac{AB}{AC}=\frac{2}{3}$ , so $AB=2\sqrt{5}$ and $AC=3\sqrt{5}$ . Now observe that $\angle{BAD}=90^{\circ}$ . Let the intersection of $BD$ and $AP$ be $X$ . Then $\angle{ABD}=90^{\circ}-\angle{BAX}=\angle{APB}$ . Consequently, \[\bigtriangleup DAB \sim \bigtriangleup ABP\] and therefore $\frac{DA}{AB} = \frac{AB}{BP}$ , so $AD=\boxed{10}$ , and we're done!


Answer: 717.0

Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$ . The tangents to $\omega$ at $B$ and $C$ intersect at $T$ . Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$ , respectively. Suppose $BT = CT = 16$ $BC = 22$ , and $TX^2 + TY^2 + XY^2 = 1143$ . Find $XY^2$

Solution:
Let $O$ be the circumcenter of $\triangle ABC$ ; say $OT$ intersects $BC$ at $M$ ; draw segments $XM$ , and $YM$ .  We have $MT=3\sqrt{15}$
Since $\angle A=\angle CBT=\angle BCT$ , we have $\cos A=\tfrac{11}{16}$ . Notice that $AXTY$ is cyclic, so $\angle XTY=180^{\circ}-A$ , so $\cos XTY=-\cos A$ , and the cosine law in $\triangle TXY$ gives \[1143-2XY^2=-\frac{11}{8}\cdot XT\cdot YT.\]
Since $\triangle BMT \cong \triangle CMT$ , we have $TM\perp BC$ , and therefore quadrilaterals $BXTM$ and $CYTM$ are cyclic. Let $P$ (resp. $Q$ ) be the midpoint of $BT$ (resp. $CT$ ). So $P$ (resp. $Q$ ) is the center of $(BXTM)$ (resp. $CYTM$ ). Then $\theta=\angle ABC=\angle MTX$ and $\phi=\angle ACB=\angle YTM$ . So $\angle XPM=2\theta$ , so \[\frac{\frac{XM}{2}}{XP}=\sin \theta,\] which yields $XM=2XP\sin \theta=BT(=CT)\sin \theta=TY$ . Similarly we have $YM=XT$
Ptolemy's theorem in $BXTM$ gives \[16TY=11TX+3\sqrt{15}BX,\] while Pythagoras' theorem gives $BX^2+XT^2=16^2$ . Similarly, Ptolemy's theorem in $YTMC$ gives \[16TX=11TY+3\sqrt{15}CY\] while Pythagoras' theorem in $\triangle CYT$ gives $CY^2+YT^2=16^2$ . Solve this for $XT$ and $TY$ and substitute into the equation about $\cos XTY$ to obtain the result $XY^2=\boxed{717}$

Solution: Let $M$ denote the midpoint of $\overline{BC}$ . The critical claim is that $M$ is the orthocenter of $\triangle AXY$ , which has the circle with diameter $\overline{AT}$ as its circumcircle. To see this, note that because $\angle BXT = \angle BMT = 90^\circ$ , the quadrilateral $MBXT$ is cyclic, it follows that \[\angle MXA = \angle MXB = \angle MTB = 90^\circ - \angle TBM = 90^\circ - \angle A,\] implying that $\overline{MX} \perp \overline{AC}$ . Similarly, $\overline{MY} \perp \overline{AB}$ . In particular, $MXTY$ is a parallelogram.  Hence, by the Parallelogram Law, \[TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).\] But $TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135$ . Therefore \[XY^2 = \frac13(2 \cdot 1143-135) = \boxed{717}.\]

Solution:
Let $H$ be the orthocenter of $\triangle AXY$
Lemma 1: $H$ is the midpoint of $BC$
Proof: Let $H'$ be the midpoint of $BC$ , and observe that $XBH'T$ and $TH'CY$ are cyclical. Define $H'Y \cap BA=E$ and $H'X \cap AC=F$ , then note that: \[\angle H'BT=\angle H'CT=\angle H'XT=\angle H'YT=\angle A.\] That implies that $\angle H'XB=\angle H'YC=90^\circ-\angle A$ $\angle CH'Y=\angle EH'B=90^\circ-\angle B$ , and $\angle BH'Y=\angle FH'C=90^\circ-\angle C$ . Thus $YH'\perp AX$ and $XH' \perp AY$ $H'$ is indeed the same as $H$ , and we have proved lemma 1.
Since $AXTY$ is cyclical, $\angle XTY=\angle XHY$ and this implies that $XHYT$ is a paralelogram.
By the Law of Cosines: \[XY^2=XT^2+TY^2+2(XT)(TY)\cdot \cos(\angle A)\] \[XY^2=XH^2+HY^2+2(XH)(HY) \cdot \cos(\angle A)\] \[HT^2=HX^2+XT^2-2(HX)(XT) \cdot \cos(\angle A)\] \[HT^2=HY^2+YT^2-2(HY)(YT) \cdot \cos(\angle A).\] We add all these equations to get: \[HT^2+XY^2=2(XT^2+TY^2) \qquad (1).\] We have that $BH=HC=11$ and $BT=TC=16$ using our midpoints. Note that $HT \perp BC$ , so by the Pythagorean Theorem, it follows that $HT^2=135$ . We were also given that $XT^2+TY^2=1143-XY^2$ , which we multiply by $2$ to use equation $(1)$ \[2(XT^2+TY^2)=2286-2 \cdot XY^2\] Since $2(XT^2+TY^2)=2(HT^2+TY^2)=HT^2+XY^2$ , we have \[135+XY^2=2286-2 \cdot XY^2\] \[3 \cdot XY^2=2151.\] Therefore, $XY^2=\boxed{717}$ . ~ MathLuis

Solution:
Using the Claim (below) we get $\triangle ABC \sim \triangle XTM \sim \triangle YMT.$
Corresponding sides of similar $\triangle XTM \sim \triangle YMT$ is $MT,$ so
$\triangle XTM = \triangle YMT \implies MY = XT, MX = TY \implies XMYT$ – parallelogram.
\[4 TD^2 = MT^2 = \sqrt{BT^2 - BM^2}  =\sqrt{153}.\] The formula for median $DT$ of triangle $XYT$ is \[2 DT^2 = XT^2 + TY^2 – \frac{XY^2}{2},\] \[3 \cdot XY^2 = 2XT^2 + 2TY^2 + 2XY^2 –  4 DT^2,\] \[3 \cdot XY^2  = 2 \cdot 1143-153 = 2151 \implies XY^2 = \boxed{717}.\]


Answer: 58.0

Let $\triangle ABC$ be an acute triangle with circumcircle $\omega,$ and let $H$ be the intersection of the altitudes of $\triangle ABC.$ Suppose the tangent to the circumcircle of $\triangle HBC$ at $H$ intersects $\omega$ at points $X$ and $Y$ with $HA=3,HX=2,$ and $HY=6.$ The area of $\triangle ABC$ can be written in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$

Solution:
The following is a power of a point solution to this menace of a problem:
 Let points be what they appear as in the diagram below. Note that $3HX = HY$ is not insignificant; from here, we set $XH = HE = \frac{1}{2} EY = HL = 2$ by PoP and trivial construction. Now, $D$ is the reflection of $A$ over $H$ . Note $AO \perp XY$ , and therefore by Pythagorean theorem we have $AE = XD = \sqrt{5}$ . Consider $HD = 3$ . We have that $\triangle HXD \cong HLK$ , and therefore we are ready to PoP with respect to $(BHC)$ . Setting $BL = x, LC = y$ , we obtain $xy = 10$ by PoP on $(ABC)$ , and furthermore, we have $KH^2 = 9 = (KL - x)(KL + y) = (\sqrt{5} - x)(\sqrt{5} + y)$ . Now, we get $4 = \sqrt{5}(y - x) - xy$ , and from $xy = 10$ we take \[\frac{14}{\sqrt{5}} = y - x.\] However, squaring and manipulating with $xy = 10$ yields that $(x + y)^2 = \frac{396}{5}$ and from here, since $AL = 5$ we get the area to be $3\sqrt{55} \implies \boxed{058}$ . ~awang11's sol

Solution:
 Diagram not to scale.
We first observe that $H'$ , the image of the reflection of $H$ over line $BC$ , lies on circle $O$ .  This is because $\angle HBC = 90 - \angle C = \angle H'AC = \angle H'BC$ .  This is a well known lemma.  The result of this observation is that circle $O'$ , the circumcircle of $\triangle BHC$ is the image of circle $O$ over line $BC$ , which in turn implies that $\overline{AH} = \overline{OO'}$ and thus $AHO'O$ is a parallelogram.  That $AHO'O$ is a parallelogram implies that $AO$ is perpendicular to $\overline{XY}$ , and thus divides segment $\overline{XY}$ in two equal pieces, $\overline{XD}$ and $\overline{DY}$ , of length $4$
Using Power of a Point, \[\overline{AH} \cdot \overline{HH'} = \overline{XH} \cdot \overline{HY} \Longrightarrow 3 \cdot \overline{HH'} = 2 \cdot 6 \Longrightarrow \overline{HH'} = 4\] This means that $\overline{HL} = \frac12 \cdot 4 = 2$ and $\overline{AL} = 2 + 3 = 5$ , where $L$ is the foot of the altitude from $A$ onto $BC$ .  All that remains to be found is the length of segment $\overline{BC}$
Looking at right triangle $\triangle AHD$ , we find that \[\overline{AD} = \sqrt{\overline{AH}^2 - \overline{HD}^2} = \sqrt{3^2 - 2^2} = \sqrt{5}\] Looking at right triangle $\triangle ODY$ , we get the equation \[\overline{OY}^2 - \overline{DY}^2 = \overline{OD}^2 = \left(\overline{AO} - \overline{AD}\right)^2\] Plugging in known values, and letting $R$ be the radius of the circle, we find that \[R^2 - 16 = (R - \sqrt{5})^2 = R^2 - 2\sqrt5 R + 5 \Longrightarrow R = \frac{21\sqrt5}{10}\]
Recall that $AHO'O$ is a parallelogram, so $\overline{AH} = \overline{OO'} = 3$ .  So, $\overline{OM} = \frac32$ , where $M$ is the midpoint of $\overline{BC}$ .  This means that \[\overline{BC} = 2\overline{BM} = 2\sqrt{R^2 - \left(\frac32\right)^2} = 2\sqrt{\frac{441}{20} - \frac{9}{4}} = \frac{6\sqrt{55}}{5}\]
Thus, the area of triangle $\triangle ABC$ is \[\frac{\overline{AL} \cdot \overline{BC}}{2} = \frac{5 \cdot \frac{6\sqrt{55}}{5}}{2} = {3\sqrt{55}}\] The answer is $3 + 55 = \boxed{058}$

Solution: Note that $\overline{BC}$ bisects $\overline{OO'}$ . Using the same method from Solution 3 we find $R = \frac{21}{2\sqrt{5}}$ . Let the midpoint of $\overline{BC}$ be $N$ , then by the Pythagorean Theorem we have $BN^2 = OB^2 - ON^2 = R^2 - \frac{3}{2}^2$ , so $BN = \frac{3\sqrt{11}}{\sqrt{5}}$ . Since $h_a = 5$ we have that the area of ABC is $3\sqrt{55}$ so the answer is $3 + 55 = \boxed{58}$


Answer: 75.0

Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\angle AEP).$

Solution:
By Miquel's theorem, $P=(AEF)\cap(BFD)\cap(CDE)$ (intersection of circles). The law of cosines can be used to compute $DE=42$ $EF=35$ , and $FD=13$ . Toss the points on the coordinate plane; let $B=(-7, 0)$ $D=(0, 0)$ , and $C=(48, 0)$ , where we will find $\tan^{2}\left(\measuredangle CDP\right)$ with $P=(BFD)\cap(CDE)$
By the extended law of sines, the radius of circle $(BFD)$ is $\frac{13}{2\sin 60^{\circ}}=\frac{13}{3}\sqrt{3}$ . Its center lies on the line $x=-\frac{7}{2}$ , and the origin is a point on it, so $y=\frac{23}{6}\sqrt{3}$
The radius of circle $(CDE)$ is $\frac{42}{2\sin 60^{\circ}}=14\sqrt{3}$ . The origin is also a point on it, and its center is on the line $x=24$ , so $y=2\sqrt{3}$
The equations of the two circles are \begin{align*}(BFD)&:\left(x+\tfrac{7}{2}\right)^{2}+\left(y-\tfrac{23}{6}\sqrt{3}\right)^{2}=\tfrac{169}{3} \\ (CDE)&:\left(x-24\right)^{2}+\left(y-2\sqrt{3}\right)^{2}=588\end{align*} These equations simplify to \begin{align*}(BFD)&:x^{2}+7x+y^{2}-\tfrac{23}{3}\sqrt{3}y=0 \\ (CDE)&: x^{2}-48x+y^{2}-4\sqrt{3}y=0\end{align*} Subtracting these two equations gives that both their points of intersection, $D$ and $P$ , lie on the line $55x-\tfrac{11}{3}\sqrt{3}y=0$ . Hence, $\tan^{2}\left(\measuredangle AEP\right)=\tan^{2}\left(\measuredangle CDP\right)=\left(\frac{55}{\tfrac{11}{3}\sqrt{3}}\right)^{2}=3\left(\tfrac{55}{11}\right)^{2}=\boxed{075}$ . To scale, the configuration looks like the figure below:

Solution:
Denote $\theta = \angle AEP$
In $AFPE$ , we have $\overrightarrow{AF} + \overrightarrow{FP} + \overrightarrow{PE} + \overrightarrow{EA} = 0$ .
Thus, \[ AF + FP e^{i \theta} + PE e^{i \left( \theta + 60^\circ \right)} + EA e^{- i 120^\circ} = 0. \]
Taking the real and imaginary parts, we get \begin{align*} AF + FP \cos \theta + PE \cos \left( \theta + 60^\circ \right) + EA \cos \left( - 120^\circ \right) & = 0 \hspace{1cm} (1) \\ FP \sin \theta + PE \sin \left( \theta + 60^\circ \right) + EA \sin \left( - 120^\circ \right) & = 0 \hspace{1cm} (2) \end{align*}
In $BDPF$ , analogous to the analysis of $AFPE$ above, we get \begin{align*} BD + DP \cos \theta + PF \cos \left( \theta + 60^\circ \right) + FB \cos \left( - 120^\circ \right) & = 0 \hspace{1cm} (3) \\ DP \sin \theta + PF \sin \left( \theta + 60^\circ \right) + FB \sin \left( - 120^\circ \right) & = 0 \hspace{1cm} (4) \end{align*}
Taking $(1) \cdot \sin \left( \theta + 60^\circ \right) - (2) \cdot \cos \left( \theta + 60^\circ \right)$ , we get \[ AF \sin \left( \theta + 60^\circ \right) + \frac{\sqrt{3}}{2} FP - EA \sin \theta = 0 . \hspace{1cm} (5) \]
Taking $(3) \cdot \sin \theta - (4) \cdot \cos \theta$ , we get \[ BD \sin \theta - \frac{\sqrt{3}}{2} FP + FB \sin \left( \theta + 120^\circ \right) . \hspace{1cm} (6) \]
Taking $(5) + (6)$ , we get \[ AF \sin \left( \theta + 60^\circ \right)  - EA \sin \theta + BD \sin \theta  + FB \sin \left( \theta + 120^\circ \right) . \]
Therefore, \begin{align*} \tan \theta & = \frac{\frac{\sqrt{3}}{2} \left( AF + FB \right)} {\frac{FB}{2} + EA - \frac{AF}{2} - BD} \\ & = 5 \sqrt{3} . \end{align*}
Therefore, $\tan^2 \theta = \boxed{075}$

Solution:
Drop the perpendiculars from $P$ to $\overline{AB}$ $\overline{AC}$ $\overline{BC}$ , and call them $Q,R,$ and $S$ respectively. This gives us three similar right triangles $FQP$ $ERP$ , and $DSP.$
The sum of the perpendiculars to a point $P$ within an equilateral triangle is always constant, so we have that $PQ+PR+PS=\dfrac{55 \sqrt{3}}{2}.$
The sum of the lengths of the alternating segments split by the perpendiculars from a point $P$ within an equilateral triangle is always equal to half the perimeter, so $QA+RC+SB = \dfrac{165}{2},$ which means that $FQ+ER+DS = QA+RC+SB - CE - AF - BD = \dfrac{165}{2} - 30 - 40 - 7 = \dfrac{11}{2}.$
Finally, $\tan AEP = \dfrac{PQ}{FQ} = \dfrac{PR}{ER} = \dfrac{PS}{DS} = \dfrac{PQ+PR+PS}{FQ+ER+DS} = 5 \sqrt{3}.$
Thus, $\tan^2 AEP = \boxed{075}.$

Solution:
This solution is inspired by AIME 1999 Problem 14 Solution 2 (a similar question)
Draw line segments from $P$ to points $A$ $B$ , and $C$ . And label the angle measure of $\angle{BFP}$ $\angle{CDP}$ , and $\angle{AEP}$ to be $\alpha$
Using Law of Cosines (note that $\cos{\angle{AFP}}=\cos{\angle{BDP}}=\cos{\angle{CEP}}=\cos{180^\circ-\alpha}=-\cos{\alpha}$
\begin{align*} (1) \ BP^2 &= FP^2+15^2-2\cdot FP\cdot15\cdot\cos(\alpha)\\ (2) \ BP^2 &= DP^2+7^2+2\cdot DP\cdot7\cdot\cos(\alpha)\\ (3) \ CP^2 &= DP^2+48^2-2\cdot DP\cdot48\cdot\cos(\alpha)\\ (4) \ CP^2 &= EP^2+30^2+2\cdot EP\cdot30\cdot\cos(\alpha)\\ (5) \ AP^2 &= EP^2+25^2-2\cdot EP\cdot25\cdot\cos(\alpha)\\ (6) \ AP^2 &= FP^2+40^2+2\cdot FP\cdot40\cdot\cos(\alpha)\\ \end{align*}
We can perform this operation $(1) - (2) + (3) - (4) + (5) - (6)$
Leaving us with (after combining and simplifying) \[\cos{\alpha}=\frac{-11}{2\cdot(DP+EP+FP)}\]
Therefore, we want to solve for $DP+EP+FP$
Notice that $\angle{DPE}=\angle{EPF}=\angle{FPD}=120^\circ$
We can use Law of Cosines again to solve for the sides of $\triangle{DEF}$ , which have side lengths of $13$ $42$ , and $35$ , and area $120\sqrt{3}$
Label the lengths of $PD$ $PE$ , and $PF$ to be $x$ $y$ , and $z$
Therefore, using the $\sin$ area formula,
\begin{align*} [\triangle{DEF}] &= \frac{1}{2}\cdot\sin{120°}\cdot(xy+yz+zx) = 120\sqrt{3} \\ xy+yz+zx &= 2^5\cdot3\cdot5 \end{align*}
In addition, we know that
\begin{align*} x^2+y^2+xy&=42^2\\ y^2+z^2+yz&=35^2\\ z^2+x^2+zx&=13^2\\ \end{align*}
By using Law of Cosines for $\triangle{DPE}$ $\triangle{EPF}$ , and $\triangle{FPD}$ respectively
Because we want $DP+EP+FP$ , which is $x+y+z$ , we see that
\begin{align*} (x+y+z)^2 &= \frac{x^2+y^2+x+y^2+z^2+yz+z^2+x^2+3(zx+xy+yz+zx)}{2} \\ (x+y+z)^2 &= \frac{42^2+35^2+13^2+3\cdot2^5\cdot3\cdot5}{2} \\ (x+y+z)^2 &= 2299 \\ x+y+z &= 11\sqrt{19} \end{align*}
So plugging the results back into the equation before, we get
\begin{align*} \cos{\alpha} &= \frac{-1}{2\sqrt{19}}\\ \sin{\alpha} &= \frac{5\sqrt{3}}{2\sqrt{19}} \end{align*}
Giving us \[\tan^2{\alpha}=\boxed{075}\]

Solution:
We begin by using the fact stated in Solution 3 that, for any point in an equilateral triangle, the lengths of the three perpendicular lines dropped to the sides of the triangle add up to the altitude of that triangle. To make things simple, let's assign $\angle AEP = \angle BFP = \angle CDP = \alpha$ . We can label these three perpendiculars as: \[PD\cdot\sin{\alpha} + PE\cdot\sin{\alpha} + PF\cdot\sin{\alpha} = \dfrac{55 \sqrt{3}}{2}\] Simplifying, we get \[PD + PE + PF = \dfrac{55 \sqrt{3}}{2\cdot\sin{\alpha}}\] Now, as stated and quoting Solution 4,
"Draw line segments from $P$ to points $A$ $B$ , and $C$ . [We know that] the angle measure of $\angle{AEP}$ $\angle{BFP}$ , and $\angle{CDP}$ is $\alpha$
Using Law of Cosines (note that $\cos{\angle{AFP}}=\cos{\angle{BDP}}=\cos{\angle{CEP}}=\cos{180^\circ-\alpha}=-\cos{\alpha}$
\begin{align*} (1) \ BP^2 &= FP^2+15^2-2\cdot FP\cdot15\cdot\cos(\alpha)\\ (2) \ BP^2 &= DP^2+7^2+2\cdot DP\cdot7\cdot\cos(\alpha)\\ (3) \ CP^2 &= DP^2+48^2-2\cdot DP\cdot48\cdot\cos(\alpha)\\ (4) \ CP^2 &= EP^2+30^2+2\cdot EP\cdot30\cdot\cos(\alpha)\\ (5) \ AP^2 &= EP^2+25^2-2\cdot EP\cdot25\cdot\cos(\alpha)\\ (6) \ AP^2 &= FP^2+40^2+2\cdot FP\cdot40\cdot\cos(\alpha)\\ \end{align*}
We can perform this operation $(1) - (2) + (3) - (4) + (5) - (6)$
Leaving us with (after combining and simplifying) \[PD + PE + PF=\frac{11}{2\cdot\cos{\alpha}}\] ".
Now, we can use our previous equation along with this one to get: \[\frac{11}{2\cdot\cos{\alpha}} = \dfrac{55 \sqrt{3}}{2\cdot\sin{\alpha}}\]
This equation becomes: \[\tan{\alpha} = 5\sqrt{3}\] As so, our answer is \[\left(5\sqrt3\right)^2=\boxed{075}.\] ~Solution by armang32324 (Mathemagics Club)

Solution:
By the law of cosines, \[FE=\sqrt{AF^2+AE^2-2AF\cdot AE\cos\angle FAE}=35.\] Similarly we get $FD=13$ and $DE=42$ $\angle AEP=\angle BFP=\angle CDP\overset\triangle=\theta$ implies that $AFPE$ $BDPF$ , and $CDPE$ are three cyclic quadrilaterals, as shown below:  Using the law of sines in each, \[\frac{AP}{35}=\frac{AP}{FE}=\frac{CP}{42}=\frac{CP}{ED}=\frac{BP}{13}=\frac{BP}{DF}=\frac{\sin\theta}{\sin\frac\pi3}.\] So we can set $AP=35k$ $BP=13k$ , and $CP=42k$ . Let $PD=d$ $PE=e$ , and $PF=f$ . Applying Ptolemy theorem in the cyclic quadrilaterals, \[\begin{cases}AP\cdot FE=AF\cdot PE+AE\cdot PF,\\CP\cdot ED=CE\cdot PD+CD\cdot PE,\\BP\cdot DF=BD\cdot PF+BF\cdot PD.\end{cases} \implies \begin{cases} 1225k=40e+25f,\\1764k=30d+48e,\\169k=15d+7f, \end{cases}\] We can solve out $d=\frac{54k}5$ $e=30k$ $f=k$ . By the law of cosines in $\triangle PEF$ $FE=\sqrt{900k^2+k^2-60k\cdot\left(\frac{-1}2\right)}=\sqrt{931}k$ . The law of sines yield $\frac{\sin\angle AEP}{\sin\angle FAE}=\frac{AP}{FE}=\frac{35k}{\sqrt{931}k}=\frac{35}{\sqrt{931}}$ .
Lastly, $\sin\angle AEP=\frac{5\sqrt{57}}{38}$ , then $\tan\angle AEP=5\sqrt3$ . The answer is \[\left(5\sqrt3\right)^2=\boxed{075}.\]


Answer: 110.0

Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$ . What is the degree measure of $\angle BFC ?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$

Solution:
Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find $\angle ABC=70^{\circ}$ . We can find $\angle ECB=20^{\circ}$ and $\angle DBC=50^{\circ}$ by the triangle angle sum on $\triangle ECB$ and $\triangle DBC$
\[\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC=180^{\circ}\implies\angle DBC=50^{\circ}\]
\[\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB=180^{\circ}\implies\angle ECB=20^{\circ}\]
Then, we take triangle $BFC$ , and find $\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{110}.$

Solution:
Alternatively, we could have used similar triangles. We start similarly to Solution 1.
Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Therefore, \[\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.\]
So, $\triangle BEF \sim BDA$ by AA Similarity, since $\angle EBF = \angle DBA$ and $\angle BEC = 90^{\circ} = \angle BDA$ .
Thus, we know \[\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.\]
Finally, we deduce \[\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{110}.\]

Solution:
Through the property of angles formed by intersecting chords, we find that \[m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}\]
Through the Outside Angles Theorem, we find that \[m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}\]
Adding the two equations gives us \[m\angle BFC + m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB\]
Since $\overarc{BC}$ is the diameter, $m\overarc{BC}=180^{\circ}$ , and because $\triangle ABC$ is isosceles and $m\angle ACB=40^{\circ}$ , we have $m\angle CAB=70^{\circ}$ . Thus \[m\angle BFC=180^{\circ}-70^{\circ}=\boxed{110}\]

Solution: Notice that if $\angle BEC = 90^{\circ}$ , then $\angle BCE$ and $\angle ACE$ must be $20^{\circ}$ . Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that $\angle EBD \cong \angle ECD = 20^{\circ}$ . Thus $\angle CBF = 70 - 20 = 50^{\circ}$ , and so $\angle BFC = 180 - 20 - 50 = 110^{\circ}$ , which is $\boxed{110}$


Answer: 110.0

Solution: $\triangle{ABC}$ is isosceles so $\angle{CAB}=70^{\circ}$ . Since $CB$ is a diameter, $\angle{CDB}=\angle{CEB}=90^{\circ}$ . Quadrilateral $ADFE$ is cyclic since $\angle{ADF}+\angle{AEF}=180^{\circ}$ . Therefore $\angle{BFC}=\angle{DFE}=180^{\circ}-\angle{CAB}=\boxed{110}$


Answer: 250.0

Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$

Solution:
This solution refers to the Diagram section.
Let $\angle PAB = \angle PBC = \angle PCA = \theta,$ from which $\angle PAC = 90^\circ-\theta,$ and $\angle APC = 90^\circ.$
Moreover, we have $\angle PBA = \angle PCB = 45^\circ-\theta,$ as shown below:  Note that $\triangle PAB \sim \triangle PBC$ by AA Similarity. The ratio of similitude is $\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},$ so $\frac{10}{PB} = \frac{1}{\sqrt2}$ and thus $PB=10\sqrt2.$ Similarly, we can figure out that $PC=20$
Finally, $AC=\sqrt{10^2+20^2}=10\sqrt{5}$ , so the area of $\triangle ABC$ is \[\frac12\cdot AB\cdot AC = \frac12\cdot (10\sqrt{5})^2 = \boxed{250}.\]

Solution:
Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$
Let the common angle be $\theta$ . Note that $\angle PAC = 90^\circ-\theta$ , thus $\angle APC = 90^\circ$ . From there, we know that $AC = \frac{10}{\sin\theta}$
Note that $\angle ABP = 45^\circ-\theta$ , so from law of sines we have \[\frac{10}{\sin\theta \cdot \frac{\sqrt{2}}{2}}=\frac{10}{\sin(45^\circ-\theta)}.\] Dividing by $10$ and multiplying across yields \[\sqrt{2}\sin(45^\circ-\theta)=\sin\theta.\] From here use the sine subtraction formula, and solve for $\sin\theta$ \begin{align*} \cos\theta-\sin\theta&=\sin\theta \\ 2\sin\theta&=\cos\theta \\ 4\sin^2\theta&=\cos^2\theta \\ 4\sin^2\theta&=1-\sin^2\theta \\ 5\sin^2\theta&=1 \\ \sin\theta&=\frac{1}{\sqrt{5}}. \end{align*} Substitute this to find that $AC=10\sqrt{5}$ , thus the area is $\frac{(10\sqrt{5})^2}{2}=\boxed{250}$

Solution:
Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$
Do some angle chasing yielding:
We have $AC=\frac{10}{\sin\theta}$ since $\triangle APC$ is a right triangle. Since $\triangle ABC$ is a $45^\circ$ $45^\circ$ $90^\circ$ triangle, $AB=\frac{10}{\sin\theta}$ , and $BC=\frac{10\sqrt{2}}{\sin\theta}$
Note that $\triangle APB \sim \triangle BPC$ by a factor of $\sqrt{2}$ . Thus, $BP = 10\sqrt{2}$ , and $PC = 20$
From Pythagorean theorem, $AC=10\sqrt{5}$ so the area of $\triangle ABC$ is $\frac{(10\sqrt{5})^2}{2}=\boxed{250}$

Solution:
Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$
Notice that in triangle $PBC$ $\angle PBC + 45-\angle PCA = 45^\circ$ , so $\angle BPC = 135^\circ$ . Similar logic shows $\angle APC = 135^\circ$
Now, we see that $\triangle APB \sim \triangle BPC$ with ratio $1:\sqrt{2}$ (as $\triangle ABC$ is a $45^\circ$ $45^\circ$ $90^\circ$ triangle). Hence, $\overline{PB}=10\sqrt{2}$ . We use the Law of Cosines to find $AB$ \begin{align*} AB^2&=BP^2+AP^2-2ab\cos(APB) \\ &=100+200-2(10)(10\sqrt{2}\cos(135^\circ)) \\ &=300+200\cdot\sqrt{2}\cdot\frac{1}{\sqrt{2}} \\ &=500. \end{align*} Since $\triangle ABC$ is a right triangle, the area is $\frac{AB^2}{2}=\frac{500}{2}=\boxed{250}$

Solution: Denote the area of $X$ by $[X].$ As in previous solutions, we see that $\angle APC = 90 ^\circ, \triangle BPC \sim \triangle APB$ with ratio $k = \sqrt{2}\implies$ \[\frac {PC}{PB} = \frac {PB}{PA} = k \implies PC = k^2 \cdot AP = 20 \implies [APC] = \frac {AP \cdot PC}{2} = 100.\] \[[BPC] = k^2 [APB] = 2 [APB].\] \[AB = BC, \angle PCA = \angle PAB \implies \frac {[APC]}{[APB]} = \frac {PC}{PA} = 2 \implies\] \[[ABC] = [APB] + [APC] + [BPC] = [APC] \cdot (\frac {1}{2} + 1 + 2 \cdot \frac {1}{2}) = \frac {5}{2} \cdot  [APC] = \boxed{250}.\] vladimir.shelomovskii@gmail.com, vvsss

Solution: Denote $\angle PCA = \theta$ . Then, by trig Ceva's: \begin{align*} \frac{\sin^3(\theta)}{\sin(90-\theta) \cdot \left(\sin(45-\theta)\right)^2} &= 1 \\ \sin^3(\theta) &= \cos(\theta) \cdot \left(\sin(45) \cos(\theta) - \cos(45) \sin(\theta)\right)^2 \\ 2\sin^3(\theta) &= \cos(\theta) \cdot \left(\cos(\theta) - \sin(\theta)\right)^2 \\ 2\sin^2(\theta) &= \cot(\theta) \cdot \left(1 - 2\sin(\theta)\cos(\theta)\right) \\ 2\sin^2(\theta) &= \cot(\theta) - 2\cos^2(\theta) \\ \cot(\theta) &= 2 \\ \sin(\theta) &= \frac{\sqrt{5}}{5}. \end{align*} Note that $\angle APC$ is a right angle. Therefore: \begin{align*} \sin(\theta) &= \frac{AP}{AC} \\ AC &= \frac{10}{\frac{\sqrt{5}}{5}} \\ &= 10\sqrt{5} \\ |ABC| &= \frac{AC^2}{2} \\ &= \boxed{250} ~ConcaveTriangle

Solution: Notice that point $P$ is one of the two Brocard Points of $\triangle ABC$ . (The angle equalities given in the problem are equivalent to the definition of a Brocard point.) By the Brocard point formula, \begin{align*} \cot(\phi) = \cot(A)+\cot(B)+\cot(C)  \end{align*} , where $\phi$ is equal to $\angle PAB$ .(This is also called the Brocard angle of triangle ABC). Because the triangle is an isosceles right triangle, the cotangents are easy to compute: \begin{align*}\cot(\phi) = 0 + 1 + 1 \\ \cot(\phi) = 2\end{align*} By definition, $\cot(\phi) = \frac{\cos(\phi)}{\sin(\phi)}$ . By the Pythagorean identity, $\cos(\phi)=\frac{2\sqrt{5}}{5}$ and $\sin(\phi) = \cos(\phi)=\frac{\sqrt{5}}{5}$ . Consider triangle $APB$ . By the problem condition, $\angle PBA = 45-\phi$ , so $\angle BPA = 135^{\circ}$ \begin{align*}\sin{45-\theta} = \sin{45}\cos{\phi}-\cos{45}\sin{\phi} = \frac{\sqrt{10}}{10}\end{align*} Now, we can use the Law of Sines. \begin{align*} \frac{AP}{\sin{45-\theta}}&=\frac{AB}{\sin{135}} \\ 10 \sqrt{10} &= \sqrt{2} AB \\ AB &= 10 \sqrt{5} \end{align*} Therefore, the answer is \[[ABC] = \frac{1}{2} (AB)^2 = \boxed{250}.\] ~ewei12


Answer: 468.0

Let $\triangle ABC$ have circumcenter $O$ and incenter $I$ with $\overline{IA}\perp\overline{OI}$ , circumradius $13$ , and inradius $6$ . Find $AB\cdot AC$

Solution: By Euler's formula $OI^{2}=R(R-2r)$ , we have $OI^{2}=13(13-12)=13$ . Thus, by the Pythagorean theorem, $AI^{2}=13^{2}-13=156$ . Let $AI\cap(ABC)=M$ ; notice $\triangle AOM$ is isosceles and $\overline{OI}\perp\overline{AM}$ which is enough to imply that $I$ is the midpoint of $\overline{AM}$ , and $M$ itself is the midpoint of $II_{a}$ where $I_{a}$ is the $A$ -excenter of $\triangle ABC$ . Therefore, $AI=IM=MI_{a}=\sqrt{156}$ and \[AB\cdot AC=AI\cdot AI_{a}=3\cdot AI^{2}=\boxed{468}.\]

Solution:
Denote $AB=a, AC=b, BC=c$ . By the given condition, $\frac{abc}{4A}=13; \frac{2A}{a+b+c}=6$ , where $A$ is the area of $\triangle{ABC}$
Moreover, since $OI\bot AI$ , the second intersection of the line $AI$ and $(ABC)$ is the reflection of $A$ about $I$ , denote that as $D$ . By the incenter-excenter lemma, $DI=BD=CD=\frac{AD}{2}\implies BD(a+b)=2BD\cdot c\implies a+b=2c$
Thus, we have $\frac{2A}{a+b+c}=\frac{2A}{3c}=6, A=9c$ . Now, we have $\frac{abc}{4A}=\frac{abc}{36c}=\frac{ab}{36}=13\implies ab=\boxed{468}$

Solution:
Denote by $R$ and $r$ the circumradius and inradius, respectively.
First, we have \[ r = 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \hspace{1cm} (1) \]
Second, because $AI \perp IO$ ,
\begin{align*}
AI & = AO \cos \angle IAO \\
& = AO \cos \left( 90^\circ - C - \frac{A}{2} \right) \\
& = AO \sin \left( C + \frac{A}{2} \right) \\
& = R \sin \left( C + \frac{180^\circ - B - C}{2} \right) \\
& = R \cos \frac{B - C}{2} .
\end{align*}
Thus,
\begin{align*}
r & = AI \sin \frac{A}{2} \\
& = R \sin \frac{A}{2} \cos \frac{B-C}{2} \hspace{1cm} (2)
\end{align*}
Taking $(1) - (2)$ , we get
\[
4 \sin \frac{B}{2} \sin \frac{C}{2} = \cos \frac{B-C}{2} .
\]
We have
\begin{align*}
2 \sin \frac{B}{2} \sin \frac{C}{2}
& = - \cos \frac{B+C}{2} + \cos \frac{B-C}{2} .
\end{align*}
Plugging this into the above equation, we get
\[
\cos \frac{B-C}{2} = 2 \cos \frac{B+C}{2} . \hspace{1cm} (3)
\]
Now, we analyze Equation (2). We have
\begin{align*}
\frac{r}{R} & = \sin \frac{A}{2} \cos \frac{B-C}{2} \\
& = \sin \frac{180^\circ - B - C}{2} \cos \frac{B-C}{2} \\
& = \cos \frac{B+C}{2} \cos \frac{B-C}{2} \hspace{1cm} (4)
\end{align*}
Solving Equations (3) and (4), we get
\[
\cos \frac{B+C}{2} = \sqrt{\frac{r}{2R}}, \hspace{1cm}
\cos \frac{B-C}{2} = \sqrt{\frac{2r}{R}} . \hspace{1cm} (5)
\]
Now, we compute $AB \cdot AC$ . We have
\begin{align*}
AB \cdot AC & = 2R \sin C \cdot 2R \sin B \\
& = 2 R^2 \left( - \cos \left( B + C \right) + \cos \left( B - C \right) \right) \\
& = 2 R^2 \left( - \left( 2 \left( \cos \frac{B+C}{2} \right)^2 - 1 \right)
+ \left( 2 \left( \cos \frac{B-C}{2} \right)^2 - 1 \right) \right) \\
& = 6 R r \\
& = \boxed{468}
\end{align*}
where the first equality follows from the law of sines, the fourth equality follows from (5).


Answer: 126.0

Let $\triangle ABC$ have side lengths $AB=30$ $BC=32$ , and $AC=34$ . Point $X$ lies in the interior of $\overline{BC}$ , and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX$ , respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}$

Solution: First note that \[\angle I_1AI_2 = \angle I_1AX + \angle XAI_2 = \frac{\angle BAX}2 + \frac{\angle CAX}2 = \frac{\angle A}2\] is a constant not depending on $X$ , so by $[AI_1I_2] = \tfrac12(AI_1)(AI_2)\sin\angle I_1AI_2$ it suffices to minimize $(AI_1)(AI_2)$ .  Let $a = BC$ $b = AC$ $c = AB$ , and $\alpha = \angle AXB$ .  Remark that \[\angle AI_1B = 180^\circ - (\angle I_1AB + \angle I_1BA) = 180^\circ - \tfrac12(180^\circ - \alpha) = 90^\circ + \tfrac\alpha 2.\] Applying the Law of Sines to $\triangle ABI_1$ gives \[\frac{AI_1}{AB} = \frac{\sin\angle ABI_1}{\sin\angle AI_1B}\qquad\Rightarrow\qquad AI_1 = \frac{c\sin\frac B2}{\cos\frac\alpha 2}.\] Analogously one can derive $AI_2 = \tfrac{b\sin\frac C2}{\sin\frac\alpha 2}$ , and so \[[AI_1I_2] = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{2\cos\frac\alpha 2\sin\frac\alpha 2} = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{\sin\alpha}\geq bc\sin\frac A2 \sin\frac B2\sin\frac C2,\] with equality when $\alpha = 90^\circ$ , that is, when $X$ is the foot of the perpendicular from $A$ to $\overline{BC}$ .  In this case the desired area is $bc\sin\tfrac A2\sin\tfrac B2\sin\tfrac C2$ .  To make this feasible to compute, note that \[\sin\frac A2=\sqrt{\frac{1-\cos A}2}=\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}2} = \sqrt{\dfrac{(a-b+c)(a+b-c)}{4bc}}.\] Applying similar logic to $\sin \tfrac B2$ and $\sin\tfrac C2$ and simplifying yields a final answer of \begin{align*}bc\sin\frac A2\sin\frac B2\sin\frac C2&=bc\cdot\dfrac{(a-b+c)(b-c+a)(c-a+b)}{8abc}\\&=\dfrac{(30-32+34)(32-34+30)(34-30+32)}{8\cdot 32}=\boxed{126}

Solution: It's clear that $\angle I_{1}AI_{2}=\frac{1}{2}\angle BAX+\frac{1}{2}\angle CAX=\frac{1}{2}\angle A$ . Thus \begin{align*} AI_{1}I_{2}&=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\angle I_{1}AI_{2} \\ &=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\left(\frac{1}{2}\angle A\right).\end{align*} By the Law of Sines on $\triangle AI_{1}B$ \[\frac{AI_{1}}{\sin\left(\frac{1}{2}\angle B\right)}=\frac{AB}{\sin\angle AI_{1}B}.\] Similarly, \[\frac{AI_{2}}{\sin\left(\frac{1}{2}\angle C\right)}=\frac{AC}{\sin\angle AI_{2}C}.\] It is well known that \[\angle AI_{1}B=90+\frac{1}{2}\angle AXB~~~\text{and}~~~\angle AI_{2}C=90+\frac{1}{2}\angle AXC.\] Denote $\alpha=\frac{1}{2}\angle AXB$ and $\theta=\frac{1}{2}\angle AXC$ , with $\alpha+\theta=90^{\circ}$ . Thus $\sin\alpha=\cos\theta$ and \begin{align*}\frac{AI_{1}}{\sin\left(\frac{1}{2}\angle B\right)}=\frac{AB}{\sin\angle AI_{1}B}\Longrightarrow\frac{AB}{\sin\left(90^{\circ}+\alpha\right)}\Longrightarrow\frac{AB}{\cos\alpha} \\\frac{AI_{2}}{\sin\left(\frac{1}{2}\angle C\right)}=\frac{AC}{\sin\angle AI_{2}C}\Longrightarrow\frac{AC}{\sin\left(90^{\circ}+\theta\right)}\Longrightarrow\frac{AC}{\cos\theta}\Longrightarrow\frac{AC}{\sin\alpha}.\end{align*} Thus \[AI_{1}=\frac{AB\sin\left(\frac{1}{2}\angle B\right)}{\cos\alpha}~~~\text{and}~~~AI_{2}=\frac{AC\sin\left(\frac{1}{2}\angle C\right)}{\sin\alpha}\] so \begin{align*}[AI_{1}I_{2}]&=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\left(\frac{1}{2}\angle A\right) \\ &=\frac{AB\sin\left(\frac{1}{2}\angle B\right)\cdot AC\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right)}{2\sin\alpha\cos\alpha} \\ &=\frac{AB\sin\left(\frac{1}{2}\angle B\right)\cdot AC\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right)}{\sin(2\alpha)}.\end{align*} We intend to minimize this expression, which is equivalent to maximizing $\sin(2\alpha)$ , and that occurs when $\alpha=45^{\circ}$ , or $\angle AXB=90^{\circ}$ . Ergo, $X$ is the foot of the altitude from $A$ to $\overline{BC}$ . In that case, we intend to compute \[AB\cdot AC\cdot\sin\left(\frac{1}{2}\angle B\right)\cdot\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right).\] Recall that \[\sin\left(\frac{1}{2}\angle B\right)=\sqrt{\frac{1-\cos\angle B}{2}}\] and similarly for angles $C$ and $A$ . Applying the Law of Cosines to each angle of $\triangle ABC$ gives \begin{align*}\angle B&:\cos\angle B=\frac{30^{2}+32^{2}-34^{2}}{2\cdot 30\cdot 32}=\frac{2}{5} \\ \angle C&:\cos\angle C=\frac{32^{2}+34^{2}-30^{2}}{2\cdot 32\cdot 34}=\frac{10}{17} \\ \angle A&:\cos\angle A=\frac{30^{2}+34^{2}-32^{2}}{2\cdot 30\cdot 34}=\frac{43}{85}.\end{align*} Thus \begin{align*}\sin\left(\frac{1}{2}\angle B\right)=\sqrt{\frac{1-\tfrac{2}{5}}{2}}=\sqrt{\frac{3}{10}} \\ \sin\left(\frac{1}{2}\angle C\right)=\sqrt{\frac{1-\tfrac{10}{17}}{2}}=\sqrt{\frac{7}{34}} \\ \sin\left(\frac{1}{2}\angle A\right)=\sqrt{\frac{1-\tfrac{43}{85}}{2}}=\sqrt{\frac{21}{85}}.\end{align*} Thus the answer is \begin{align*} & 30\cdot 34\cdot\sqrt{\frac{3}{10}\cdot\frac{7}{34}\cdot\frac{21}{85}} \\ =&~30\cdot 34\cdot\sqrt{\frac{3}{2\cdot 5}\cdot\frac{7}{2\cdot 17}\cdot\frac{3\cdot 7}{5\cdot 17}} \\ =&~30\cdot 34\cdot\frac{3\cdot 7}{2\cdot 5\cdot 17} \\ =&~(2\cdot 3\cdot 5)\cdot(2\cdot 17)\cdot\frac{3\cdot 7}{2\cdot 5\cdot 17} \\ =&~2\cdot 3^{2}\cdot 7 \\ =&~\boxed{126}

Solution:
First, instead of using angles to find $[AI_1I_2]$ , let's try to find the area of other, simpler figures, and subtract that from $[ABC]$ . However, to do this, we need to be able to figure out the length of the inradii, and so, we need to find $AX$
To minimize $[AI_1I_2]$ , intuitively, we should try to minimize the length of $AX$ , since, after using the $rs=A$ formula for the area of a triangle, we'll be able to minimize the inradii lengths, and thus, eventually minimize the area of $[AI_1I_2]$ . (Proof needed here).
We need to minimize $AX$ . Let $AX=d$ $BX=s$ , and $CX=32-s$ . After an application of Stewart's Theorem, we will get that \[d=\sqrt{s^2-24s+900}\] To minimize this quadratic, $s=12$ whereby we conclude that $d=6\sqrt{21}$
From here, draw perpendiculars down from $I_1$ and $I_2$ to $AB$ and $AC$ respectively, and label the foot of these perpendiculars $D$ and $E$ respectively. After, draw the inradii from $I_1$ to $BX$ , and from $I_2$ to $CX$ , and draw in $I_1I_2$
Label the foot of the inradii to $BX$ and $CX$ $F$ and $G$ , respectively. From here, we see that to find $[AI_1I_2]$ , we need to find $[ABC]$ , and subtract off the sum of $[DBCEI_2I_1], [ADI_1],$ and $[AEI_2]$
$[DBCEI_2I_1]$ can be found by finding the area of two quadrilaterals $[DBFI_1]+[ECGI_2]$ as well as the area of a trapezoid $[FGI_2I_1]$ . If we let the inradius of $ABX$ be $r_1$ and if we let the inradius of $ACX$ be $r_2$ , we'll find, after an application of basic geometry and careful calculations on paper, that $[DBCEI_2I_1]=13r_1+19r_2$
The area of two triangles can be found in a similar fashion, however, we must use $XYZ$ substitution to solve for $AD$ as well as $AE$ . After doing this, we'll get a similar sum in terms of $r_1$ and $r_2$ for the area of those two triangles which is equal to \[\frac{(9+3\sqrt{21})(r_1)}{2} + \frac{(7+3\sqrt{21})(r_2)}{2}.\]
Now we're set. Summing up the area of the Hexagon and the two triangles and simplifying, we get that the formula for $[AI_1I_2]$ is just \[[ABC]-\left(\frac{(35+3\sqrt{21})(r_1)}{2}+\frac{(45+3r_2\sqrt{21})(r_2)}{2}\right).\]
Using Heron's formula, $[ABC]=96\sqrt{21}$ . Solving for $r_1$ and $r_2$ using Heron's in $ABX$ and $ACX$ , we get that $r_1=3\sqrt{21}-9$ and $r_2=3\sqrt{21}-7$ . From here, we just have to plug into our above equation and solve.
Doing so gets us that the minimum area of $AI_1I_2=\boxed{126}.$


Answer: 15.0

Let $\triangle A_0B_0C_0$ be a triangle whose angle measures are exactly $59.999^\circ$ $60^\circ$ , and $60.001^\circ$ . For each positive integer $n$ , define $A_n$ to be the foot of the altitude from $A_{n-1}$ to line $B_{n-1}C_{n-1}$ . Likewise, define $B_n$ to be the foot of the altitude from $B_{n-1}$ to line $A_{n-1}C_{n-1}$ , and $C_n$ to be the foot of the altitude from $C_{n-1}$ to line $A_{n-1}B_{n-1}$ . What is the least positive integer $n$ for which $\triangle A_nB_nC_n$ is obtuse?
$\textbf{(A) } 10 \qquad \textbf{(B) }11 \qquad \textbf{(C) } 13\qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$

Solution:
For all nonnegative integers $n$ , let $\angle C_nA_nB_n=x_n$ $\angle A_nB_nC_n=y_n$ , and $\angle B_nC_nA_n=z_n$
Note that quadrilateral $A_0B_0A_1B_1$ is cyclic since $\angle A_0A_1B_0=\angle A_0B_1B_0=90^\circ$ ; thus, $\angle A_0A_1B_1=\angle A_0B_0B_1=90^\circ-x_0$ . By a similar argument, $\angle A_0A_1C_1=\angle A_0C_0C_1=90^\circ-x_0$ . Thus, $x_1=\angle A_0A_1B_1+\angle A_0A_1C_1=180^\circ-2x_0$ . By a similar argument, $y_1=180^\circ-2y_0$ and $z_1=180^\circ-2z_0$
Therefore, for any positive integer $n$ , we have $x_n=180^\circ-2x_{n-1}$ (identical recurrence relations can be derived for $y_n$ and $z_n$ ). To find an explicit form for this recurrence, we guess that the constant term is related exponentially to $n$ (and the coefficient of $x_0$ is $(-2)^n$ ). Hence, we let $x_n=pq^n+r+(-2)^nx_0$ . We will solve for $p$ $q$ , and $r$ by iterating the recurrence to obtain $x_1=180^\circ-2x_0$ $x_2=4x_0-180^\circ$ , and $x_3=540-8x_0$ . Letting $n=1,2,3$ respectively, we have \begin{align} pq+r&=180 \\ pq^2+r&=-180 \\ pq^3+r&=540 \end{align}
Subtracting $(1)$ from $(3)$ , we have $pq(q^2-1)=360$ , and subtracting $(1)$ from $(2)$ gives $pq(q-1)=-360$ . Dividing these two equations gives $q+1=-1$ , so $q=-2$ . Substituting back, we get $p=-60$ and $r=60$
We will now prove that for all positive integers $n$ $x_n=-60(-2)^n+60+(-2)^nx_0=(-2)^n(x_0-60)+60$ via induction. Clearly the base case of $n=1$ holds, so it is left to prove that $x_{n+1}=(-2)^{n+1}(x_0-60)+60$ assuming our inductive hypothesis holds for $n$ . Using the recurrence relation, we have \begin{align*} x_{n+1}&=180-2x_n \\ &=180-2((-2)^n(x_0-60)+60) \\ &=(-2)^{n+1}(x_0-60)+60 \end{align*}
Our induction is complete, so for all positive integers $n$ $x_n=(-2)^n(x_0-60)+60$ . Identical equalities hold for $y_n$ and $z_n$
The problem asks for the smallest $n$ such that either $x_n$ $y_n$ , or $z_n$ is greater than $90^\circ$ . WLOG, let $x_0=60^\circ$ $y_0=59.999^\circ$ , and $z_0=60.001^\circ$ . Thus, $x_n=60^\circ$ for all $n$ $y_n=-(-2)^n(0.001)+60$ , and $z_n=(-2)^n(0.001)+60$ . Solving for the smallest possible value of $n$ in each sequence, we find that $n=15$ gives $y_n>90^\circ$ . Therefore, the answer is $\boxed{15}$

Solution: We note that the problem seems quite complicated, but since it is an AMC 12, the difference between the largest angle of $\triangle A_nB_nC_n$ and $60^{\circ}$ (we call this quantity S) most likely reduces to a simpler problem like some repeating sequence. The only obvious sequence (for the answer choices) is a geometric sequence with an integer common ratio. From here, we get that $S_n=0.001S^n$ and we need $S_n>30$ . The first power of two greater than $30,000$ is $2^{15}$ thus our answer is $\boxed{15}$


Answer: 26.0

Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY = 90^{\circ}$ . Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$ , respectively. Given that $XN = 19$ and $YM = 22$ , find $XY$
$\mathrm{(A)}\ 24 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 28 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 32$

Solution:
2002 12B AMC-20.png
Let $OM = x$ $ON = y$ . By the Pythagorean Theorem on $\triangle XON, MOY$ respectively, \begin{align*} (2x)^2 + y^2 &= 19^2\\ x^2 + (2y)^2 &= 22^2\end{align*}
Summing these gives $5x^2 + 5y^2 = 845 \Longrightarrow x^2 + y^2 = 169$
By the Pythagorean Theorem again, we have
\[(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = \sqrt{4(169)} = \sqrt{676} = \boxed{26}\]


Answer: 26.0

Solution:
Let $XO=x$ and $YO=y.$ Then, $XY=\sqrt{x^2+y^2}.$
Since $XN=19$ and $YM=22,$ \[XN^2=19^2=x^2+(\dfrac{y}{2})^2)=\dfrac{x^2}{4}+y^2\] \[YM^2=22^2=(\dfrac{x}{2})^2+y^2=\dfrac{x^2}{4}+y^2.\]
Adding these up: \[19^2+22^2=\dfrac{4x^2+y^2}{4}+\dfrac{x^2+4y^2}{4}\] \[845=\dfrac{5x^2+5y^2}{4}\] \[3380=5x^2+5y^2\] \[676=x^2+y^2.\]
Then, we substitute: $XY=\sqrt{x^2+y^2}=\sqrt{676}=\boxed{26}.$


Answer: 26.0

Solution:
2002 12B AMC-20.png
Let $OM = x$ $ON = y$ . By the Pythagorean Theorem on $\triangle XON, MOY$ respectively, \begin{align*} (2x)^2 + y^2 &= 19^2\\ x^2 + (2y)^2 &= 22^2\end{align*}
Summing these gives $5x^2 + 5y^2 = 845 \Longrightarrow x^2 + y^2 = 169$
By the Pythagorean Theorem again, we have
\[(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = \sqrt{4(169)} = \sqrt{676} = \boxed{26}\]


Answer: 26.0

Solution:
Let $XO=x$ and $YO=y.$ Then, $XY=\sqrt{x^2+y^2}.$
Since $XN=19$ and $YM=22,$ \[XN^2=19^2=x^2+(\dfrac{y}{2})^2)=\dfrac{x^2}{4}+y^2\] \[YM^2=22^2=(\dfrac{x}{2})^2+y^2=\dfrac{x^2}{4}+y^2.\]
Adding these up: \[19^2+22^2=\dfrac{4x^2+y^2}{4}+\dfrac{x^2+4y^2}{4}\] \[845=\dfrac{5x^2+5y^2}{4}\] \[3380=5x^2+5y^2\] \[676=x^2+y^2.\]
Then, we substitute: $XY=\sqrt{x^2+y^2}=\sqrt{676}=\boxed{26}.$


Answer: 725.0

Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle . Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$

Solution:
Let $P = (0,0)$ be at the origin. Using the formula $A = rs$ on $\triangle PQR$ , where $r_{1}$ is the inradius (similarly define $r_2, r_3$ to be the radii of $C_2, C_3$ ), $s = \frac{PQ + QR + RP}{2} = 180$ is the semiperimeter , and $A = \frac 12 bh = 5400$ is the area, we find $r_{1} = \frac As = 30$ . Or, the inradius could be directly by using the formula $\frac{a+b-c}{2}$ , where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. (This formula should be used only for right triangles .) Thus $ST, UV$ lie respectively on the lines $y = 60, x = 60$ , and so $RS = 60, UQ = 30$
Note that $\triangle PQR \sim \triangle STR \sim \triangle UQV$ . Since the ratio of corresponding lengths of similar figures are the same, we have
\[\frac{r_{1}}{PR} = \frac{r_{2}}{RS} \Longrightarrow r_{2} = 15\ \text{and} \ \frac{r_{1}}{PQ} = \frac{r_{3}}{UQ} \Longrightarrow r_{3} = 10.\]
Let the centers of $\odot C_2, C_3$ be $O_2 = (0 + r_{2}, 60 + r_{2}) = (15, 75), O_3 = (60 + r_{3}, 0 + r_{3}) = (70,10)$ , respectively; then by the distance formula we have $O_2O_3 = \sqrt{55^2 + 65^2} = \sqrt{10 \cdot 725}$ . Therefore, the answer is $n = \boxed{725}$

Solution:
We compute $r_1 = 30, r_2 = 15, r_3 = 10$ as above. Let $A_1, A_2, A_3$ respectively the points of tangency of $C_1, C_2, C_3$ with $QR$
By the Two Tangent Theorem , we find that $A_{1}Q = 60$ $A_{1}R = 90$ . Using the similar triangles, $RA_{2} = 45$ $QA_{3} = 20$ , so $A_{2}A_{3} = QR - RA_2 - QA_3 = 85$ . Thus $(O_{2}O_{3})^{2} = (15 - 10)^{2} + (85)^{2} = 7250\implies n=\boxed{725}$

Solution:
The radius of an incircle is $r=A_t/\text{semiperimeter}$ . The area of the triangle is equal to $\frac{90\times120}{2} = 5400$ and the semiperimeter is equal to $\frac{90+120+150}{2} = 180$ . The radius, therefore, is equal to $\frac{5400}{180} = 30$ . Thus using similar triangles the dimensions of the triangle circumscribing the circle with center $C_2$ are equal to $120-2(30) = 60$ $\frac{1}{2}(90) = 45$ , and $\frac{1}{2}\times150 = 75$ . The radius of the circle inscribed in this triangle with dimensions $45\times60\times75$ is found using the formula mentioned at the very beginning. The radius of the incircle is equal to $15$
Defining $P$ as $(0,0)$ $C_2$ is equal to $(60+15,15)$ or $(75,15)$ . Also using similar triangles, the dimensions of the triangle circumscribing the circle with center $C_3$ are equal to $90-2(30)$ $\frac{1}{3}\times120$ $\frac{1}{3}\times150$ or $30,40,50$ . The radius of $C_3$ by using the formula mentioned at the beginning  is $10$ . Using $P$ as $(0,0)$ $C_3$ is equal to $(10, 60+10)$ or $(10,70)$ . Using the distance formula, the distance between $C_2$ and $C_3$ $\sqrt{(75-10)^2 +(15-70)^2}$ this equals $\sqrt{7250}$ or $\sqrt{725\times10}$ , thus $n$ is $\boxed{725}$

Solution: We can calculate the inradius of each triangle as with the previous solutions. Now, notice that the hexagon $PSTVU$ is a square with its corner cut off. We literally complete the square and mark the last corner as point X. Now, construct right triangle with $C_3C_2$ as its hypotenuse. The right angle will be at point $Y$ . We will now find the length of each leg and use the Pythagorean Theorem to compute the desired length. We see that the length of $C_3Y$ is $50 + 15 = 65$ , as seen by the inradius of $C_2$ and $10$ less than the square's side length. $C_2Y$ is $45 + 10 = 55$ , which is $15$ less than the square plus the inradius of $C_3$ . Our final answer is $\sqrt{65^2 + 55^2} = \sqrt{7250} = \sqrt{(10)(725)} \rightarrow \boxed{725}$