Answer: 20.0

In $\triangle ABC$ , the sides have integer lengths and $AB=AC$ . Circle $\omega$ has its center at the incenter of $\triangle ABC$ . An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$ , and the other two excircles are both externally tangent to $\omega$ . Find the minimum possible value of the perimeter of $\triangle ABC$

Solution:
Let the tangent circle be $\omega$ . Some notation first: let $BC=a$ $AB=b$ $s$ be the semiperimeter, $\theta=\angle ABC$ , and $r$ be the inradius. Intuition tells us that the radius of $\omega$ is $r+\frac{2rs}{s-a}$ (using the exradius formula). However, the sum of the radius of $\omega$ and $\frac{rs}{s-b}$ is equivalent to the distance between the incenter and the the $B/C$ excenter. Denote the B excenter as $I_B$ and the incenter as $I$ .
Lemma: $I_BI=\frac{2b*IB}{a}$ We draw the circumcircle of $\triangle ABC$ . Let the angle bisector of $\angle ABC$ hit the circumcircle at a second point $M$ . By the incenter-excenter lemma, $AM=CM=IM$ . Let this distance be $\alpha$ . Ptolemy's theorem on $ABCM$ gives us \[a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}\] Again, by the incenter-excenter lemma, $II_B=2IM$ so $II_b=\frac{2b*IB}{a}$ as desired.
Using this gives us the following equation: \[\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}\] Motivated by the $s-a$ and $s-b$ , we make the following substitution: $x=s-a, y=s-b$ This changes things quite a bit. Here's what we can get from it: \[a=2y, b=x+y, s=x+2y\] It is known (easily proved with Heron's and $a=rs$ ) that \[r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}\] Using this, we can also find $IB$ : let the midpoint of $BC$ be $N$ . Using Pythagorean's Theorem on $\triangle INB$ \[IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y}\] We now look at the RHS of the main equation: \[r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}\] Cancelling some terms, we have \[\frac{r(x+4y)}{x}=IB\] Squaring, \[\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)\] Expanding and moving terms around gives \[(x-8y)(x+2y)=0\to x=8y\] Reverse substituting, \[s-a=8s-8b\to b=\frac{9}{2}a\] Clearly the smallest solution is $a=2$ and $b=9$ , so our answer is $2+9+9=\boxed{020}$ -franchester

Solution:
First, assume $BC=2$ and $AB=AC=x$ . The triangle can be scaled later if necessary. Let $I$ be the incenter and let $r$ be the inradius. Let the points at which the incircle intersects $AB$ $BC$ , and $CA$ be denoted $M$ $N$ , and $O$ , respectively.
Next, we calculate $r$ in terms of $x$ . Note the right triangle formed by $A$ $I$ , and $M$ . The length $IM$ is equal to $r$ . Using the Pythagorean Theorem, the length $AN$ is $\sqrt{x^2-1}$ , so the length $AI$ is $\sqrt{x^2-1}-r$ . Note that $BN$ is half of $BC=2$ , and by symmetry caused by the incircle, $BN=BM$ and $BM=1$ , so $MA=x-1$ . Applying the Pythagorean Theorem to $AIM$ , we get \[r^2+(x-1)^2=\left(\sqrt{x^2-1}-r\right)^2.\] Expanding yields \[r^2+x^2-2x+1=x^2-1-2r\sqrt{x^2-1}+r^2,\] which can be simplified to \[2r\sqrt{x^2-1}=2x-2.\] Dividing by $2$ and then squaring results in \[r^2(x^2-1)=(x-1)^2,\] and isolating $r^2$ gets us \[r^2=\frac{(x-1)^2}{x^2-1}=\frac{(x-1)^2}{(x+1)(x-1)}=\frac{x-1}{x+1},\] so $r=\sqrt{\frac{x-1}{x+1}}$
We then calculate the radius of the excircle tangent to $BC$ . We denote the center of the excircle $E_N$ and the radius $r_N$
Consider the quadrilateral formed by $M$ $I$ $E_N$ , and the point at which the excircle intersects the extension of $AB$ , which we denote $H$ . By symmetry caused by the excircle, $BN=BH$ , so $BH=1$
Note that triangles $MBI$ and $NBI$ are congruent, and $HBE$ and $NBE$ are also congruent. Denoting the measure of angles $MBI$ and $NBI$ measure $\alpha$ and the measure of angles $HBE$ and $NBE$ measure $\beta$ , straight angle $MBH=2\alpha+2\beta$ , so $\alpha + \beta=90^\circ$ . This means that angle $IBE$ is a right angle, so it forms a right triangle.
Setting the base of the right triangle to $IE$ , the height is $BN=1$ and the base consists of $IN=r$ and $EN=r_N$ . Triangles $INB$ and $BNE$ are similar to $IBE$ , so $\frac{IN}{BN}=\frac{BN}{EN}$ , or $\frac{r}{1}=\frac{1}{r_N}$ . This makes $r_N$ the reciprocal of $r$ , so $r_N=\sqrt{\frac{x+1}{x-1}}$
Circle $\omega$ 's radius can be expressed by the distance from the incenter $I$ to the bottom of the excircle with center $E_N$ . This length is equal to $r+2r_N$ , or $\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}$ . Denote this value $r_\omega$
Finally, we calculate the distance from the incenter $I$ to the closest point on the excircle tangent to $AB$ , which forms another radius of circle $\omega$ and is equal to $r_\omega$ . We denote the center of the excircle $E_M$ and the radius $r_M$ . We also denote the points where the excircle intersects $AB$ and the extension of $BC$ using $J$ and $K$ , respectively. In order to calculate the distance, we must find the distance between $I$ and $E_M$ and subtract off the radius $r_M$
We first must calculate the radius of the excircle. Because the excircle is tangent to both $AB$ and the extension of $AC$ , its center must lie on the angle bisector formed by the two lines, which is parallel to $BC$ . This means that the distance from $E_M$ to $K$ is equal to the length of $AN$ , so the radius is also $\sqrt{x^2-1}$
Next, we find the length of $IE_M$ . We can do this by forming the right triangle $IAE_M$ . The length of leg $AI$ is equal to $AN$ minus $r$ , or $\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}$ . In order to calculate the length of leg $AE_M$ , note that right triangles $AJE_M$ and $BNA$ are congruent, as $JE_M$ and $NA$ share a length of $\sqrt{x^2-1}$ , and angles $E_MAJ$ and $NAB$ add up to the right angle $NAE_M$ . This means that $AE_M=BA=x$
Using Pythagorean Theorem, we get \[IE_M=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}.\] Bringing back \[r_\omega=IE_M-r_M\] and substituting in some values, the equation becomes \[r_\omega=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}-\sqrt{x^2-1}.\] Rearranging and squaring both sides gets \[\left(r_\omega+\sqrt{x^2-1}\right)^2=\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2.\] Distributing both sides yields \[r_\omega^2+2r_\omega\sqrt{x^2-1}+x^2-1=x^2-1-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.\] Canceling terms results in \[r_\omega^2+2r_\omega\sqrt{x^2-1}=-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.\] Since \[-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}=-2\sqrt{(x+1)(x-1)\frac{x-1}{x+1}}=-2(x-1),\] We can further simplify to \[r_\omega^2+2r_\omega\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2.\] Substituting out $r_\omega$ gets \[\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)^2+2\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2\] which when distributed yields \[\frac{x-1}{x+1}+4+4\left(\frac{x+1}{x-1}\right)+2(x-1+2(x+1))=-2(x-1)+\frac{x-1}{x+1}+x^2.\] After some canceling, distributing, and rearranging, we obtain \[4\left(\frac{x+1}{x-1}\right)=x^2-8x-4.\] Multiplying both sides by $x-1$ results in \[4x+4=x^3-x^2-8x^2+8x-4x+4,\] which can be rearranged into \[x^3-9x^2=0\] and factored into \[x^2(x-9)=0.\] This means that $x$ equals $0$ or $9$ , and since a side length of $0$ cannot exist, $x=9$
As a result, the triangle must have sides in the ratio of $9:2:9$ . Since the triangle must have integer side lengths, and these values share no common factors greater than $1$ , the triangle with the smallest possible perimeter under these restrictions has a perimeter of $9+2+9=\boxed{020}$ . ~ emerald_block

Solution:
Before we start thinking about the problem, let’s draw it out;

For the sake of space, I've drawn only 2 of the 3 excircles because the third one looks the same as the second large one because the triangle is isosceles. By the incenter-excenter lemma, $AII_A$ and $BII_B$ are collinear, $E$ is the tangent of circle $I_B$ to $AC$ $F$ is the tangent of that circle to the extension of $BC$ , and $J$ is the tangent of the circle to the extension of $BA$ . The interesting part of the diagram is circle $\omega$ , which is internally tangent to circle $I_A$ yet externally tangent to circle $I_B$ . Therefore, perhaps we can relate the radius of this circle to the semiperimeter of triangle $ABC$
We can see that the radius of circle $\omega$ is $2r_{I_A}+r$ using the incenter and A-excenter of our main triangle. This radius is also equal to $BI_B - BI - r_{I_B}$ from the incenter and B-excenter of our triangle. Thus, we can solve for each of these separately in terms of the lengths of the triangle and set them equal to each other to form an equation.
To find the left hand side of the equation, we have to first find $r$ and $r_{I_A}$ . Let $a = AB = AC, b = BD = DC,$ and $h = AD$ . Then since the perimeter of the triangle is $2a+2b$ , the semiperimeter is $a+b$
Now let's take a look at triangle $BDI$ . Because $BI$ is the angle bisector of $\angle B$ , by the angle bisector theorem, $\frac{AI}{ID} = \frac{BA}{BD} \implies \frac{h-r}{r} = \frac{a}{b}$ . Rearranging, we get $r = \frac{hb}{a+b}$
Take a look at triangle $AGI$ $AG = a - GB = a - BD = a-b$ $AI = h-r = \frac{ha}{a+b}$ (angle bisector theorem), and $GI = r = \frac{hb}{a+b}$ . Now let's analyze triangle $AHI_A$ $AH = AB + BH = AB+ BD = a+b$ $AI_A = h+r_{I_A}$ , and $HI_A = r_{I_A}$ . Since $\angle GAI = \angle HAI_A$ and $\angle IGA = \angle I_AHA = 90^{\circ}$ , triangle $AGI$ and $AHI_A$ are similar by AA. Then $\frac{r_{I_A}}{r} = \frac{h+r_{I_A}}{h-r} \implies r_{I_A} = r \cdot \frac{h+r_{I_A}}{h-r} = \frac{hb}{a+b} \cdot \frac{h+r_{I_A}}{\frac{ha}{a+b}} = \frac{b(h+r_{I_A})}{a}$ . Now, solving yields $r_{I_A} = \frac{hb}{a-b}$
Finally, the left hand side of our equation is \[\frac{2hb}{a-b} + \frac{hb}{a+b}\]
Now let's look at triangle $BFI_B$ . How will we find $BI_B$ ? Let's first try to find $BF$ and $I_BF$ in terms of the lengths of the triangle. We recognize:
$BF = BC + CF = BC + CK$ . We really want to have $CA$ instead of $CK$ , and $AK$ looks very similar in length to $DC$ , so let's try to prove that they are equal.
$BJ = BF$ , so we can try to add these two and see if we get anything interesting. We have: $BJ + BF = BA + AJ + BC + CF = BA + AE + BC + CE = BA + BC + CA$ , which is our perimeter. Thus, $BF = a+b$
Triangle $BDI$ is similar to triangle $BFI_B$ by AA, and we know that $BD = b$ , and $ID = r = \frac{hb}{a+b}$ , so thus $I_BF = BF \cdot \frac{ID}{BD} = (a+b) \cdot \frac{\frac{hb}{a+b}}{b} = \frac{hb}{b} = h$ . Thus, the height of this triangle is $h$ by similarity ratios, the same height as vertex $A$ . By the Pythagorean Theorem, $BI_B = \sqrt{(a+b)^2 + h^2}$ and by similarity ratios, $BI = \frac{b}{a+b} \cdot \sqrt{(a+b)^2 + h^2}$ . Finally, $r_{I_B} = I_BF = h$ , and thus the right hand side of our equation is \[\sqrt{(a+b)^2 + h^2} - \frac{b}{a+b} \cdot \sqrt{(a+b)^2 + h^2} - h = \sqrt{(a+b)^2 + h^2}(1 - \frac{b}{a+b}) - h = \sqrt{(a+b)^2 + h^2} \cdot \frac{a}{a+b} - h\]
Setting the two equal, we have \[\frac{2hb}{a-b} + \frac{hb}{a+b} = \sqrt{(a+b)^2 + h^2} \cdot \frac{a}{a+b} - h\]
Multiplying both sides by $(a+b)(a-b)$ we have $2hb(a+b) + hb(a-b) = \sqrt{(a+b)^2 + h^2} \cdot a(a-b) - h(a^2 - b^2)$
From here, let $b = 1$ arbitrarily; note that we can always scale this value to fit the requirements later. Thus our equation is $2h(a+1) + h(a-1) = \sqrt{(a+1)^2 + h^2} \cdot a(a-1) - h(a^2 - 1)$ . Now since $h = \sqrt{a^2 - b^2}$ , we can plug into our equation:
$2h(a+1) + h(a-1) = \sqrt{a² + 2a + 1 + a^2 - b^2 } \cdot a(a-1) - h(a^2 - 1)$ . Remembering $b = 1$
$\implies 2h(a+1) + h(a-1) = \sqrt{2a^2 + 2a} \cdot a(a-1) - h(a^2 - 1)$
$\implies 3ha + h + ha^2 - h = \sqrt{2a^2 + 2a} \cdot a(a-1)$
$\implies ah(3+a) = a(a-1) \cdot \sqrt{2a^2 + 2a}$
$\implies h^2(3+a)^2 = (a-1)^2 \cdot 2a(a+1)$
$\implies (a^2 - 1) (3 + a)^2 = 2a(a+1)(a - 1)^2$
$\implies (3+ a)^2 = 2a(a-1)$
$\implies 9 + 6a + a^2 = 2a^2 - 2a$
$\implies a^2 - 8a - 9 = 0$
$\implies (a-9)(a+1) = 0$
$\implies a = 9$ because the side lengths have to be positive numbers. Furthermore, because our values for $a$ and $b$ are relatively prime, we don't have to scale down our triangle further, and we are done. Therefore, our answer is $2a + 2b = 18 + 2 = \boxed{020}$

Solution: Notice that the $A$ -excircle would have to be very small to fit the property that it is internally tangent to $\omega$ and the other two excircles are both externally tangent, given that circle $\omega$ 's centre is at the incenter of $\triangle ABC$ . If $BC=2$ , we see that $AB=AC$ must be somewhere in the $6$ to $13$ range. If we test $6$ by construction, we notice the $A$ -excircle is too big for it to be internally tangent to $\omega$ while the other two are externally tangent. This means we should test $8$ or $9$ next. I actually did this and found that $9$ worked, so the answer is $2+9+9=\boxed{20}$ . Note that $BC$ cannot be $1$ because then $AB=AC$ would have to be $4.5$ which is not an integer.

Solution:
Let $M$ be the midpoint $BC, BM = a, AB= BC = b,$ $s = b+a$ be the semiperimeter, $r$ be the inradius. Let $I_A, I_B$ be excenters, $r_A, r_B$ be exradius, $R$ be radius $\omega.$ Then $R = r + 2 r_A,$ \[r = \sqrt{\frac{(s-2a)(s-b)^2}{s}} = a \sqrt{\frac{b-a}{b+a}},\] \[r \cdot s = r_A \cdot (s-2a) \implies r_A =  a \sqrt{\frac{b+a}{b–a}},\] \[r \cdot s = r_B \cdot (s-b) \implies r_B =  \sqrt{b^2 – a^2} = AM.\] \[II_B =R + r_B = r + 2r_A + r_B \implies\] \[II_B=   a \sqrt{\frac{b-a}{b+ a}}+ 2a  \sqrt{\frac{b+a}{b – a}} + \sqrt{b^2-a^2} = b\frac{3a+b}{\sqrt{b^2-a^2}}.\] \[\overline{AI_B}= \frac{\overline{A} \cdot 2a - \overline{B} \cdot b + \overline{C}\cdot b}{2a -b + b}-\overline{A} =b \frac{\overline{C}- \overline{B}}{2a},\] \[AI_B = b, \overline{AI_B}\perp AI \implies II_B = \sqrt{AI_B^2 + (AM – r)^2},\] \[II_B =  \sqrt{b^2 +\left(\sqrt{b^2-a^2}-  a \sqrt{\frac{b-a}{b+ a}}\right)^2} = b\sqrt{\frac{2b}{b+a}}\] Therefore we get problem’s condition in the form of \[b\frac{b +3a}{\sqrt{b^2-a^2}} = b \sqrt{\frac {2b}{b+a}} \implies  b + 3a = \sqrt{2b(b-a)} \implies (b-9a)(b+a) = 0 \implies b = 9a.\]
We use $a = 1$ an get $b = 9, 2s = 18+2 = \boxed{020}$


Answer: 2.0

In $\triangle ABC$ , we have $AB = 1$ and $AC = 2$ . Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$
$\mathrm{(A)}\ \frac{1+\sqrt{2}}{2} \qquad\mathrm{(B)}\ \frac{1+\sqrt{3}}2 \qquad\mathrm{(C)}\ \sqrt{2} \qquad\mathrm{(D)}\ \frac 32 \qquad\mathrm{(E)}\ \sqrt{3}$

Solution: There is a theorem in geometry known as Pappus's Median Theorem. It states that if you have $\triangle{ABC}$ , and you draw a median from point $A$ to side $BC$ (label this as $M$ ), then: $(AM)^2 = \dfrac{2(b^2) + 2(c^2) - (a^2)}{4}$ . Note that $b$ is the length of side $\overline{AC}$ $c$ is the length of side $\overline{AB}$ , and $a$ is length of side $\overline{BC}$ . Let $MB = MC = x$ . Then $AM = 2x$ . Now, we can plug into the formula given above: $AM = 2x$ $b = 2$ $c = 1$ , and $a = 2x$ . After some simple algebra, we find $x = \dfrac{\sqrt{2}}{2}$ . Then, $BC = \boxed{2}$


Answer: 2.0

Solution: From Stewart's Theorem , we have $(2)(1/2)a(2) + (1)(1/2)a(1) = (a)(a)(a) + (1/2)a(a)(1/2)a.$ Simplifying, we get $(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{2}.$ - awu2014


Answer: 3.0

In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$
$\textbf{(A) }\ 3      \qquad \textbf{(B) }\ 2\sqrt{3}      \qquad \textbf{(C) }\ 4      \qquad \textbf{(D) }\ 5      \qquad \textbf{(E) }\ 4\sqrt{2}$

Solution: Draw height $CH$ (Perpendicular line from point C to line AD). We have that $BH=1$ . By the Pythagorean Theorem $CH=\sqrt{48}$ . Since $CD=8$ $HD=\sqrt{8^2-48}=\sqrt{16}=4$ , and $BD=HD-1$ , so $BD=\boxed{3}$

Solution: After drawing out a diagram, let $\angle{ABC}=\theta$ . By the Law of Cosines, $7^2=2^2+7^2-2(7)(2)\cos{\theta} \rightarrow 0=4-28\cos{\theta} \rightarrow \cos{\theta}=\frac{1}{7}$ . In $\triangle CBD$ , we have $\angle{CBD}=(180-\theta)$ , and using the identity $\cos(180-\theta)=-\cos{\theta}$ and Law of Cosines one more time: $8^2=7^2+x^2-2(7)(x)\left( \frac{-1}{7} \right) \rightarrow 64=49+x^2+2x \rightarrow x^2+2x-15=0$ . The only positive value for $x$ is $3$ , which gives the length of $\overline{BD}$ . Thus the answer is $\boxed{3}$

Solution:
Let $BD=k$ . Then, by Stewart's Theorem
$2k(2+k)+7^2(2+k)=7^2k+8^2\cdot 2 \implies k^2+2k-15=0 \implies k=\boxed{3}$


Answer: 3.0

Solution: Draw height $CH$ (Perpendicular line from point C to line AD). We have that $BH=1$ . By the Pythagorean Theorem $CH=\sqrt{48}$ . Since $CD=8$ $HD=\sqrt{8^2-48}=\sqrt{16}=4$ , and $BD=HD-1$ , so $BD=\boxed{3}$

Solution: After drawing out a diagram, let $\angle{ABC}=\theta$ . By the Law of Cosines, $7^2=2^2+7^2-2(7)(2)\cos{\theta} \rightarrow 0=4-28\cos{\theta} \rightarrow \cos{\theta}=\frac{1}{7}$ . In $\triangle CBD$ , we have $\angle{CBD}=(180-\theta)$ , and using the identity $\cos(180-\theta)=-\cos{\theta}$ and Law of Cosines one more time: $8^2=7^2+x^2-2(7)(x)\left( \frac{-1}{7} \right) \rightarrow 64=49+x^2+2x \rightarrow x^2+2x-15=0$ . The only positive value for $x$ is $3$ , which gives the length of $\overline{BD}$ . Thus the answer is $\boxed{3}$

Solution:
Let $BD=k$ . Then, by Stewart's Theorem
$2k(2+k)+7^2(2+k)=7^2k+8^2\cdot 2 \implies k^2+2k-15=0 \implies k=\boxed{3}$


Answer: 35.0

In $\triangle ABC$ , we have $\angle C = 3\angle A$ $a = 27$ and $c = 48$ . What is $b$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(14,0), C=(10,6); draw(A--B--C--cycle); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$a$", B--C, dir(B--C)*dir(-90)); label("$b$", A--C, dir(C--A)*dir(-90)); label("$c$", A--B, dir(A--B)*dir(-90));[/asy]
$\mathrm{(A)\ } 33 \qquad \mathrm{(B) \ }35 \qquad \mathrm{(C) \  } 37 \qquad \mathrm{(D) \  } 39 \qquad \mathrm{(E) \  }\text{not uniquely determined}$

Solution: Let $\angle A = x^{\circ}$ , so $\angle C = 3x^{\circ}$ , and thus $\angle B = \left(180-4x\right)^{\circ}$ . Now let $D$ be a point on side $AB$ such that $\angle ACD = x^{\circ}$ , so $\angle BCD = 3x^{\circ}-x^{\circ} = 2x^{\circ}$ , which gives \[\angle CDB = 180^{\circ}-2x^{\circ}-\left(180-4x\right)^{\circ} = 2x^{\circ},\] meaning that $\triangle CDB$ and $\triangle CDA$ are both isosceles, with $BC = BD$ and $AD = CD$ . In particular, $BD = BC = 27$ and $CD = AD = AB-BD = 48-27 = 21$ . Hence by Stewart's theorem on triangle $ABC$ \begin{align*}&BD \cdot AB \cdot AD + CD^2 \cdot AB = AC^2 \cdot BD + BC^2 \cdot AD \\ &\iff 27 \cdot 48 \cdot 21 + 21^2 \cdot 48 = AC^2 \cdot 27 + 27^2 \cdot 21 \\ &\iff AC^2 = \frac{27(21)(48-27) + 21^2 \cdot 48}{27} \\ &\iff AC^2 = \frac{21^2(27+48)}{27} \\ &\iff AC^2 = \frac{21^2 \cdot 25}{9} \\ &\iff AC = \frac{21 \cdot 5}{3} \qquad \text{(as } AC > 0\text{)} \\ &\iff AC = \boxed{35}

Solution:
We apply the law of sines in the form \[\frac{\sin(A)}{a} = \frac{\sin(C)}{c},\] yielding \[\frac{\sin(A)}{27} = \frac{\sin(3A)}{48} \iff 9\sin(3A) = 16\sin(A).\]
Now, the angle sum and double angle identities give \begin{align*}\sin(3A) &= \sin(2A+A) \\ &= \sin(2A)\cos(A)+\cos(2A)\sin(A) \\ &= \left(2\sin(A)\cos(A)\right)\cos(A)+\left(\cos^2(A)-\sin^2(A)\right)\sin(A) \\ &= 2\sin(A)\cos^2(A)+\sin(A)\cos^2(A)-\sin^3(A) \\ &= 3\sin(A)\left(1-\sin^2(A)\right)-\sin^3(A) \\ &\text{(using the further identity } \cos^2(\theta)+\sin^2(\theta) = 1\text{)} \\ &= 3\sin(A)-4\sin^3(A).\end{align*}
Thus our equation becomes \begin{align*}9\left(3\sin(A)-4\sin^3(A)\right) = 16\sin(A) &\iff 27\sin(A)-36\sin^3(A) = 16\sin(A) \\ &\iff 36\sin^3(A) = 11\sin(A) \\ &\iff \sin(A) = 0 \text{ or } \pm\frac{\sqrt{11}}{6}.\end{align*} Notice, however, that we must have $0^{\circ} < A < 45^{\circ}$ , the latter because otherwise $A+3A \geq 180^{\circ}$ , which would contradict the fact that $A$ and $3A$ are angles in a (non-degenerate) triangle. This means $\sin(A) > 0$ , so the only valid solution is \[\sin(A) = \frac{\sqrt{11}}{6},\] and the fact that $A$ is acute also means $\cos(A) > 0$ , so we deduce \begin{align*}\cos(A) &= \sqrt{1-\left(\frac{\sqrt{11}}{6}\right)^2} \\ &= \sqrt{1-\frac{11}{36}} \\ &=\frac{5}{6}.\end{align*} Accordingly, using the double angle identities again, \begin{align*}\sin(4A) &= \sin(2 \cdot 2A) \\ &= 2\sin(2A)\cos(2A) \\ &= 2\left(2\sin(A)\cos(A)\right)\left(\cos^2(A)-\sin^2(A)\right) \\ & =2\left(2 \cdot \frac{\sqrt{11}}{6} \cdot\frac{5}{6}\right)\left(\left(\frac{5}{6}\right)^2-\left(\frac{\sqrt{11}}{6}\right)^2\right) \\ &= \frac{5\sqrt{11}}{9} \cdot \frac{25-11}{36} \\ &= \frac{5\sqrt{11}}{9} \cdot \frac{7}{18} \\ &= \frac{35\sqrt{11}}{162}.\end{align*} Finally, the law of sines now gives \begin{align*}\frac{\sin(A)}{27} &= \frac{\sin(B)}{b} \\ &= \frac{\sin(180^{\circ}-3A-A)}{b} \\ &= \frac{\sin(4A)}{b} \qquad \text{(using the identity } \sin\left(180^{\circ}-\theta\right) = \sin(\theta)\text{)},\end{align*} so, substituting the above results, \[\frac{\left(\frac{\sqrt{11}}{6}\right)}{27} = \frac{\left(\frac{35\sqrt{11}}{162}\right)}{b} \iff b = \frac{6 \cdot 27 \cdot 35}{162} = \boxed{35}.\]


Answer: 13.0

In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Suppose we label the angles as shown below.  As $\angle BCD$ and $\angle BAD$ intercept the same arc, we know that $\angle BAD=\gamma$ . Similarly, $\angle ABD=\gamma$ . Also, using $\triangle ICA$ , we find $\angle CIA=180-\alpha-\gamma$ . Therefore, $\angle AID=\alpha+\gamma$ . Therefore, $\angle DAI=\angle AID=\alpha+\gamma$ , so $\triangle AID$ must be isosceles with $AD=ID=5$ . Similarly, $BD=ID=5$ . Then $\triangle DLB \sim \triangle ALC$ , hence $\frac{AL}{AC} = \frac{3}{5}$ . Also, $AI$ bisects $\angle LAC$ , so by the Angle Bisector Theorem $\frac{CI}{IL} =\frac{AC}{AL}= \frac{5}{3}$ . Thus $CI = \frac{10}{3}$ , and the answer is $\boxed{013}$

Solution: WLOG assume $\triangle ABC$ is isosceles. Then, $L$ is the midpoint of $AB$ , and $\angle CLB=\angle CLA=90^\circ$ . Draw the perpendicular from $I$ to $CB$ , and let it meet $CB$ at $E$ . Since $IL=2$ $IE$ is also $2$ (they are both inradii). Set $BD$ as $x$ . Then, triangles $BLD$ and $CEI$ are similar, and $\tfrac{2}{3}=\tfrac{CI}{x}$ . Thus, $CI=\tfrac{2x}{3}$ $\triangle CBD \sim \triangle CEI$ , so $\tfrac{IE}{DB}=\tfrac{CI}{CD}$ . Thus $\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}$ . Solving for $x$ , we have: $x^2-2x-15=0$ , or $x=5, -3$ $x$ is positive, so $x=5$ . As a result, $CI=\tfrac{2x}{3}=\tfrac{10}{3}$ and the answer is $\boxed{013}$

Solution: WLOG assume $\triangle ABC$ is isosceles (with vertex $C$ ). Let $O$ be the center of the circumcircle, $R$ the circumradius, and $r$ the inradius. A simple sketch will reveal that $\triangle ABC$ must be obtuse (as an acute triangle will result in $LI$ being greater than $DL$ ) and that $O$ and $I$ are collinear. Next, if $OI=d$ $DO+OI=R+d$ and $R+d=DL+LI=5$ . Euler gives us that $d^{2}=R(R-2r)$ , and in this case, $r=LI=2$ . Thus, $d=\sqrt{R^{2}-4R}$ . Solving for $d$ , we have $R+\sqrt{R^{2}-4R}=5$ , then $R^{2}-4R=25-10R+R^{2}$ , yielding $R=\frac{25}{6}$ . Next, $R+d=5$ so $d=\frac{5}{6}$ . Finally, $OC=OI+IC$ gives us $R=d+IC$ , and $IC=\frac{25}{6}-\frac{5}{6}=\frac{10}{3}$ . Our answer is then $\boxed{013}$

Solution: Since $\angle{LAD} = \angle{BDC}$ and $\angle{DLA}=\angle{DCB}$ $\triangle{DLA}\sim\triangle{DBC}$ . Also, $\angle{DAC}=\angle{BLC}$ and $\angle{ACD}=\angle{LCB}$ so $\triangle{DAC}\sim\triangle{BLC}$ . Now we can call $AC$ $b$ and $BC$ $a$ . By angle bisector theorem, $\frac{AD}{DB}=\frac{AC}{BC}$ . So let $AD=bk$ and $DB=ak$ for some value of $k$ . Now call $IC=x$ . By the similar triangles we found earlier, $\frac{3}{ak}=\frac{bk}{x+2}$ and $\frac{b}{x+5}=\frac{x+2}{a}$ . We can simplify this to $abk^2=3x+6$ and $ab=(x+5)(x+2)$ . So we can plug the $ab$ into the first equation and get $(x+5)(x+2)k^2=3(x+2) \rightarrow k^2(x+5)=3$ . We can now draw a line through $A$ and $I$ that intersects $BC$ at $E$ . By mass points, we can assign a mass of $a$ to $A$ $b$ to $B$ , and $a+b$ to $D$ . We can also assign a mass of $(a+b)k$ to $C$ by angle bisector theorem. So the ratio of $\frac{DI}{IC}=\frac{(a+b)k}{a+b}=k=\frac{2}{x}$ . So since $k=\frac{2}{x}$ , we can plug this back into the original equation to get $\left(\frac{2}{x}\right)^2(x+5)=3$ . This means that $\frac{3x^2}{4}-x-5=0$ which has roots -2 and $\frac{10}{3}$ which means our $CI=\frac{10}{3}$ and our answer is $\boxed{013}$

Solution: Since $\angle BCD$ and $\angle BAD$ both intercept arc $BD$ , it follows that $\angle BAD=\gamma$ . Note that $\angle AID=\alpha+\gamma$ by the external angle theorem. It follows that $\angle DAI=\angle AID=\alpha+\gamma$ , so we must have that $\triangle AID$ is isosceles, yielding $AD=ID=5$ . Note that $\triangle DLA \sim \triangle DAC$ , so $\frac{DA}{DL} = \frac{DC}{DA}$ . This yields $DC = \frac{25}{3}$ . It follows that $CI = DC - DI = \frac{10}{3}$ , giving a final answer of $\boxed{013}$

Solution: We can just say that quadrilateral $ADBC$ is a right kite with right angles at $A$ and $B$ . Let us construct another similar right kite with the points of tangency on $AC$ and $BC$ called $E$ and $F$ respectively, point $I$ , and point $C$ . Note that we only have to look at one half of the circle since the diagram is symmetrical. Let us call $CI$ $x$ for simplicity's sake. Based on the fact that $\triangle BCD$ is similar to $\triangle FCI$ we can use triangle proportionality to say that $BD$ is $2\frac{x+5}{x}$ . Using geometric mean theorem we can show that $BL$ must be $\sqrt{3x+6}$ . With Pythagorean Theorem we can say that $3x+6+9=4{(\frac{x+5}{x})}^2$ . Multiplying both sides by $x^2$ and moving everything to LHS will give you $3{x}^3+11{x}^2-40x-100=0$ Since $x$ must be in the form $\frac{p}{q}$ we can assume that $x$ is most likely a positive fraction in the form $\frac{p}{3}$ where $p$ is a factor of $100$ . Testing the factors in synthetic division would lead $x = \frac{10}{3}$ , giving us our desired answer $\boxed{013}$ . ~Lopkiloinm

Solution:
 Connect $D$ to $A$ and $D$ to $B$ to form quadrilateral $ACBD$ . Since quadrilateral $ACBD$ is cyclic, we can apply Ptolemy's Theorem on the quadrilateral.
Denote the length of $BD$ and $AD$ as $z$ (they must be congruent, as $\angle ABD$ and $\angle DAB$ are both inscribed in arcs that have the same degree measure due to the angle bisector intersecting the circumcircle at $D$ ), and the lengths of $BC$ $AC$ $AB$ , and $CI$ as $a,b,c, x$ , respectively.
After applying Ptolemy's, one will get that:
\[z(a+b)=c(x+5)\]
Next, since $ACBD$ is cyclic, triangles $ALD$ and $CLB$ are similar, yielding the following equation once simplifications are made to the equation $\frac{AD}{CB}=\frac{AL}{BL}$ , with the length of $BL$ written in terms of $a,b,c$ using the angle bisector theorem on triangle $ABC$
\[zc=3(a+b)\]
Next, drawing in the bisector of $\angle BAC$ to the incenter $I$ , and applying the angle bisector theorem, we have that:
\[cx=2(a+b)\]
Now, solving for $z$ in the second equation, and $x$ in the third equation and plugging them both back into the first equation, and making the substitution $w=\frac{a+b}{c}$ , we get the quadratic equation:
\[3w^2-2w-5=0\]
Solving, we get $w=5/3$ , which gives $z=5$ and $x=10/3$ , when we rewrite the above equations in terms of $w$ . Thus, our answer is $\boxed{013}$ and we're done.

Solution: Let $AB=c,BC=a,CA=b$ , and $x=\tfrac{a+b}{c}$ . Then, notice that $\tfrac{CI}{IL}=\tfrac{a+b}{c}=x$ , so $CI=IL\cdot{}x=2x$ . Also, by the incenter-excenter lemma, $AD=BD=ID=IL+LD=5$ . Therefore, by Ptolemy's Theorem on cyclic quadrilateral $ABCD$ $5a+5b=c(2x+5)$ , so $5\left(\tfrac{a+b}{c}\right)=2x+5$ , so $5x=2x+5$ . Solving, we get that $x=\tfrac{5}{3}$ , so $CI=\tfrac{10}{3}$ and the answer is $10+3=\boxed{013}$

Solution: Without loss of generality, let $\triangle ABC$ be isosceles. Note that by the incenter-excenter lemma, $DI = DA = DB.$ Hence, $DA=DB=5.$ Let the point of tangency of the incircle and $\overline{BC}$ be $F$ and the point of tangency of the incircle and $\overline{AC}$ be $E.$ We note that $\angle ALC = \angle BLC = 90^\circ$ and $LA=LB=4,$ which immediately gives $AE=BF=4.$ Applying the Pythagorean Theorem on $\triangle ALC$ and $\triangle IEC$ gives $2^2+x^2=y^2$ and $4^2+(2+y)^2 = (4+x)^2.$ Solving for $y$ gives us $y=\frac{10}{3}.$ Therefore, $IC = \frac{10}{3}$ so the answer is $\boxed{13}.$


Answer: 547.0

In $\triangle ABC$ with $AB=AC,$ point $D$ lies strictly between $A$ and $C$ on side $\overline{AC},$ and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC.$ The degree measure of $\angle ABC$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution:
If we set $\angle{BAC}$ to $x$ , we can find all other angles through these two properties:
1. Angles in a triangle sum to $180^{\circ}$ .
2. The base angles of an isosceles triangle are congruent.
Now we angle chase. $\angle{ADE}=\angle{EAD}=x$ $\angle{AED} = 180-2x$ $\angle{BED}=\angle{EBD}=2x$ $\angle{EDB} = 180-4x$ $\angle{BDC} = \angle{BCD} = 3x$ $\angle{CBD} = 180-6x$ . Since $AB = AC$ as given by the problem, $\angle{ABC} = \angle{ACB}$ , so $180-4x=3x$ . Therefore, $x = 180/7^{\circ}$ , and our desired angle is \[180-4\left(\frac{180}{7}\right) = \frac{540}{7}\] for an answer of $\boxed{547}$

Solution:
Let $\angle{BAC}$ be $x$ in degrees. $\angle{ADE}=x$ .
By Exterior Angle Theorem on triangle $AED$ $\angle{BED}=2x$ .
By Exterior Angle Theorem on triangle $ADB$ $\angle{BDC}=3x$ .
This tells us $\angle{BCA}=\angle{ABC}=3x$ and $3x+3x+x=180$ .
Thus $x=\frac{180}{7}$ and we want $\angle{ABC}=3x=\frac{540}{7}$ to get an answer of $\boxed{547}$

Solution:
graph soon
We write equations based on the triangle sum of angles theorem. There are angles that do not need variables as the less variables the better.
\begin{align*} \angle A &= y \\ \angle B &= x+z \\ \angle C &= \frac{180-x}{2} \\ \end{align*}
Then, using triangle sum of angles theorem, we find that
\begin{align*} \angle A + \angle B + \angle C = x+y+z+\frac{180-x}{2}=180 \\ \end{align*}
Now we just need to find the variables.
\begin{align*} (180-2y)+z = 180& \\ (180-2z)+y+\frac{180-x}{2} = 180& \\ \end{align*}
Notice how all the equations equal 180. We can use this to write
\begin{align*} (180-2y)+z = (180-2z)+y+\frac{180-x}{2}=x+y+z+\frac{180-x}{2} \\ \end{align*}
Simplifying, we get
\begin{align*} (180-2y)+z=(180-2z)+y+\frac{180-x}{2} \\ 360-4y+2z=360-4z+2y+180-x \\ \end{align*}
\begin{align*} 6z=6y+180-x \\ x=6y-6z+180 \\ \end{align*}
\begin{align*} (180-2y)+z=6y-6z+180+y+z+\frac{180-(6y-6+180)}{2} \\ 360-4y+2z=12y-12z+360+2y+2z+180-6y+6z-180 \\ \end{align*}
\begin{align*} 6z=12y& \\ z=2y& \\ \end{align*}
Theres more. We are at a dead end right now because we forgot that the problem states that the triangle is isosceles. With this, we can write the equation
\begin{align*} \frac{180-x}{2}=x+z \\ \end{align*}
Substituting $z$ with $2y$ , we get
\begin{align*} \frac{180-x}{2}=x+2y \\ 180-x=2x+4y \\ \end{align*} \begin{align*} 180-(6y-6z+180)=2(6y-6z+180)+4y& \\ 180-6y+12y-180=12y-24y+360+4y& \\ \end{align*} \begin{align*} 6y=-8y+360& \\ \end{align*}
With this, we get
\begin{align*} y=\frac{180}{7} \\ x=\frac{180}{7} \\ z=\frac{360}{7} \\ \end{align*}
And a final answer of $\frac{180}{7}+\frac{360}{7} = \frac{540}{7} = \boxed{547}$


Answer: 10.8

In $\triangle ABC$ with $\overline{AB}=\overline{AC}=3.6$ , a point $D$ is taken on $AB$ at a distance $1.2$ from $A$ . Point $D$ is joined to $E$ in the prolongation of $AC$ so that $\triangle AED$ is equal in area to $ABC$ . Then $\overline{AE}$ is: $\textbf{(A)}\ 4.8 \qquad\textbf{(B)}\ 5.4\qquad\textbf{(C)}\ 7.2\qquad\textbf{(D)}\ 10.8\qquad\textbf{(E)}\ 12.6$

Solution: Note that $\frac{1}{2}AB * AC *\sin\angle BAC = \frac{1}{2}AD * AE *\sin\angle DAE$ . Since $\angle BAC = \angle DAE$ , we have $AB*AC = AD*AE$ , so that $3.6*3.6 = 1.2*AE$ . Therefore, $AE = \frac{3.6^2}{1.2} = 10.8$ . Thusly, our answer is $\boxed{10.8}$ , and we are done.


Answer: 9.0

In $\triangle ABC$ with integer side lengths, $\cos A = \frac{11}{16}$ $\cos B = \frac{7}{8}$ , and $\cos C = -\frac{1}{4}$ . What is the least possible perimeter for $\triangle ABC$
$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$

Solution:
Notice that by the Law of Sines, $a:b:c = \sin{A}:\sin{B}:\sin{C}$ , so let's flip all the cosines using $\sin^{2}{x} + \cos^{2}{x} = 1$ $\sin{x}$ is positive for $0^{\circ} < x < 180^{\circ}$ , so we're good there).
$\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}$
These are in the ratio $3:2:4$ , so our minimal triangle has side lengths $2$ $3$ , and $4$ $\boxed{9}$ is our answer.

Solution: $\angle ACB$ is obtuse since its cosine is negative, so we let the foot of the altitude from $C$ to $AB$ be $H$ . Let $AH=11x$ $AC=16x$ $BH=7y$ , and $BC=8y$ . By the Pythagorean Theorem, $CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}$ and $CH=\sqrt{64y^2-49y^2}=y\sqrt{15}$ . Thus, $y=3x$ . The sides of the triangle are then $16x$ $11x+7(3x)=32x$ , and $24x$ , so for some integers $a,b$ $16x=a$ and $24x=b$ , where $a$ and $b$ are minimal. Hence, $\frac{a}{16}=\frac{b}{24}$ , or $3a=2b$ . Thus the smallest possible positive integers $a$ and $b$ that satisfy this are $a=2$ and $b=3$ , so $x=\frac{1}{8}$ . The sides of the triangle are $2$ $3$ , and $4$ , so $\boxed{9}$ is our answer.

Solution:
Using the law of cosines, we get the following equations:
\[c^2=a^2+b^2+\frac{ab}{2}\] \[b^2=a^2+c^2-\frac{7ac}{4}\] \[a^2=b^2+c^2-\frac{11bc}{8}\]
Substituting $a^2+c^2-\frac{7ac}{4}$ for $b^2$ in $a^2=b^2+c^2-\frac{11bc}{8}$ and simplifying, we get the following: \[14a+11b=16c\]
Note that since $a, b, c$ are integers, we can solve this for integers. By some trial and error, we get that $(a,b,c) = (3,2,4)$ . Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is $3+2+4 = \boxed{9}$


Answer: 4.0

In $\triangle ABC$ with right angle at $C$ , altitude $CH$ and median $CM$ trisect the right angle. If the area of $\triangle CHM$ is $K$ , then the area of $\triangle ABC$ is
$\textbf{(A)}\ 6K\qquad\textbf{(B)}\ 4\sqrt3\ K\qquad\textbf{(C)}\ 3\sqrt3\ K\qquad\textbf{(D)}\ 3K\qquad\textbf{(E)}\ 4K$

Solution:
Draw diagram as shown (note that $A$ and $B$ can be interchanged, but it doesn’t change the solution).
Note that because $CM$ is a median, $AM = BM$ .  Also, by ASA Congruency, $\triangle CHA = \triangle CHM$ , so $AH = HM$ .  That means $HM = \tfrac{1}{4} \cdot AB$ , and since $\triangle CHM$ and $\triangle ABC$ share an altitude, $[ABC] = \boxed{4}$


Answer: 247.0

In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$ can be written as $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution:
Because $M$ is the midpoint of $BC$ , following from the Stewart's theorem, $AM = 2 \sqrt{37}$
Because $A$ $B$ $C$ , and $P$ are concyclic, $\angle BPA = \angle C$ $\angle CPA = \angle B$
Denote $\theta = \angle PBQ$
In $\triangle BPQ$ , following from the law of sines, \[ \frac{BQ}{\sin \angle BPA} = \frac{PQ}{\angle PBQ} \]
Thus, \[ \frac{BQ}{\sin C} = \frac{PQ}{\sin \theta} . \hspace{1cm} (1) \]
In $\triangle CPQ$ , following from the law of sines, \[ \frac{CQ}{\sin \angle CPA} = \frac{PQ}{\angle PCQ} \]
Thus, \[ \frac{CQ}{\sin B} = \frac{PQ}{\sin \theta} . \hspace{1cm} (2) \]
Taking $\frac{(1)}{(2)}$ , we get \[ \frac{BQ}{\sin C} = \frac{CQ}{\sin B} \]
In $\triangle ABC$ , following from the law of sines, \[ \frac{AB}{\sin C} = \frac{AC}{\sin B} . \hspace{1cm} (3) \]
Thus, Equations (2) and (3) imply \begin{align*} \frac{BQ}{CQ} & = \frac{AB}{AC} \\ & = \frac{13}{15} . \hspace{1cm} (4) \end{align*}
Next, we compute $BQ$ and $CQ$
We have \begin{align*} BQ^2 & = AB^2 + AQ^2 - 2 AB\cdot AQ \cos \angle BAQ \\ & = AB^2 + AQ^2 - 2 AB\cdot AQ \cos \angle BAM \\ & = AB^2 + AQ^2 - 2 AB\cdot AQ \cdot \frac{AB^2 + AM^2 - BM^2}{2 AB \cdot AM} \\ & = AB^2 + AQ^2 -  AQ \cdot \frac{AB^2 + AM^2 - BM^2}{AM} \\ & = 169 + AQ^2 - \frac{268}{2 \sqrt{37}} AQ .  \hspace{1cm} (5) \end{align*}
We have \begin{align*} CQ^2 & = AC^2 + AQ^2 - 2 AC\cdot AQ \cos \angle CAQ \\ & = AC^2 + AQ^2 - 2 AC\cdot AQ \cos \angle CAM \\ & = AC^2 + AQ^2 - 2 AC\cdot AQ \cdot \frac{AC^2 + AM^2 - CM^2}{2 AC \cdot AM} \\ & = AC^2 + AQ^2 -  AQ \cdot \frac{AC^2 + AM^2 - CM^2}{AM} \\ & = 225 + AQ^2 - \frac{324}{2 \sqrt{37}} AQ .  \hspace{1cm} (6) \end{align*}
Taking (5) and (6) into (4), we get $AQ = \frac{99}{\sqrt{148}}$
Therefore, the answer is $99 + 148 = \boxed{247}$

Solution:
Define $L_1$ to be the foot of the altitude from $A$ to $BC$ . Furthermore, define $L_2$ to be the foot of the altitude from $Q$ to $BC$ . From here, one can find $AL_1=12$ , either using the 13-14-15 triangle or by calculating the area of $ABC$ two ways. Then, we find $BL_1=5$ and $L_1C = 9$ using Pythagorean theorem. Let $QL_2=x$ . By AA similarity, $\triangle{AL_1M}$ and $\triangle{QL_2M}$ are similar. By similarity ratios, \[\frac{AL_1}{L_1M}=\frac{QL_2}{L_2M}\] \[\frac{12}{2}=\frac{x}{L_2M}\] \[L_2M = \frac{x}{6}\] Thus, $BL_2=BM-L_2M=7-\frac{x}{6}$ . Similarly, $CL_2=7+\frac{x}{6}$ . Now, we angle chase from our requirement to obtain new information. \[\angle{PBQ}=\angle{PCQ}\] \[\angle{QCM}+\angle{PCM}=\angle{QBM}+\angle{PBM}\] \[\angle{QCL_2}+\angle{PCM}=\angle{QBL_2}+\angle{PBM}\] \[\angle{PCM}-\angle{PBM}=\angle{QBL_2}-\angle{QCL_2}\] \[\angle{MAB}-\angle{MAC}=\angle{QBL_2}-\angle{QCL_2}\text{(By inscribed angle theorem)}\] \[(\angle{MAL_1}+\angle{L_1AB})-(\angle{CAL_1}-\angle{MAL_1})=\angle{QBL_2}-\angle{QCL_2}\] \[2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1}=\angle{QBL_2}-\angle{QCL_2}\] Take the tangent of both sides to obtain \[\tan(2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1})=\tan(\angle{QBL_2}-\angle{QCL_2})\] By the definition of the tangent function on right triangles, we have $\tan{MAL_1}=\frac{7-5}{12}=\frac{1}{6}$ $\tan{CAL_1}=\frac{9}{12}=\frac{3}{4}$ , and $\tan{L_1AB}=\frac{5}{12}$ . By abusing the tangent angle addition formula, we can find that \[\tan(2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1})=\frac{196}{2397}\] By substituting $\tan{\angle{QBL_2}}=\frac{6x}{42-x}$ $\tan{\angle{QCL_2}}=\frac{6x}{42+x}$ and using tangent angle subtraction formula we find that \[x=\frac{147}{37}\] Finally, using similarity formulas, we can find \[\frac{AQ}{AM}=\frac{12-x}{x}\] . Plugging in $x=\frac{147}{37}$ and $AM=\sqrt{148}$ , we find that \[AQ=\frac{99}{\sqrt{148}}\] Thus, our final answer is $99+148=\boxed{247}$ .
~sigma

Solution:
It is clear that $BQCP$ is a parallelogram. By Stewart's Theorem, $AM=\sqrt{148}$ , POP on $M$ tells $PM=\frac{49}{\sqrt{148}}$
As $QM=PM, AQ=AM-PM=\frac{99}{\sqrt{148}}$ leads to $\boxed{247}$

Solution:
First, note that by Law of Sines, $\frac{\sin(\angle{QBP})}{QP}=\frac{\sin(\angle{BPQ})}{BQ}$ and that $\frac{\sin(\angle{QCP})}{QP}=\frac{\sin(\angle{QPC})}{QP}$ . Equating the 2 expressions, you get that $\frac{\sin(\angle{BPQ})}{BQ}=\frac{\sin(\angle{QPC})}{QP}$ . Now drop the altitude from $A$ to $BC$ . As it is commonly known that the dropped altitude forms a $5-12-13$ and a $9-12-15$ triangle, you get the measures of $\angle{ABC}$ and $\angle{ACB}$ respectively, which are $\arcsin(\frac{12}{13})$ and $\arcsin(\frac{4}{5})$ . However, by the inscribed angle theorem, you get that $\angle{BPQ}=\arcsin(\frac{4}{5})$ and that $\angle{QPC}=\arcsin(\frac{12}{13})$ , respectively. Therefore, by Law of Sines (as previously stated) $\frac{BQ}{CQ}=\frac{13}{15}$
Now commence coordbashing. Let $B$ be the origin, and $A$ be the point $(5,12)$ . As $AP$ passes through $A$ , which is $(5,12)$ , and $M$ , which is $(7,0)$ , it has the equation $-6x+42$ , so therefore a point on this line can be written as $(x,42-6x)$ . As we have the ratio of the lengths, which prompts us to write the lengths in terms of the distance formula, we can just plug and chug it in to get the ratio $\frac{\sqrt{37x^2-504x+1764}}{\sqrt{37x^2-532x+1960}}=\frac{13}{15}$ . This can be squared to get $\frac{37x^2-504x+1764}{37x^2-532x+1960}=\frac{169}{225}$ . This can be solved to get a solution of $x=\frac{469}{74}$ , and an extraneous solution of $5$ which obviously doesn’t work.
Plugging $x$ into the line equation gets you $y=\frac{147}{37}$ . The distance between this point and $A$ , which is $(5,12)$ is $\sqrt{\frac{9801}{148}}$ , or simplified to $\frac{99}{\sqrt{148}}\Longrightarrow99+148=\boxed{247}$

Solution:
We use the law of Cosine and get \[AB^2 = AM^2 + BM^2 - 2 AM \cdot BM \cos \angle AMB,\] \[AC^2 = AM^2 + CM^2 + 2 AM \cdot CM \cos \angle AMB \implies\] \[AM^2 = \frac {AB^2 + AC^2}{2}- BM^2 = \sqrt{148} \approx 12.\] We use the power of point $M$ with respect circumcircle $\triangle ABC$ and get \[AM \cdot MP = BM \cdot CM = BM^2 \implies\] \[PM = \frac {49}{\sqrt {148}} \approx \frac {48}{12} \approx 4 < AM.\] It is clear that if $Q = P,$ then $\angle PBQ = \angle PCQ = 0 \implies$
if $Q$ is symmetric to $P$ with respect $M$ then $\angle PBQ = \angle PCQ.$
There exists a unique point $Q$ on segment $\overline{AM}, PM < AM \implies$ \[PQ = AM - PM = \frac{99}{\sqrt{148}} \implies \boxed{247}.\] vladimir.shelomovskii@gmail.com, vvsss


Answer: 6.5

In $\triangle ABC, AB = 13, BC = 14$ and $CA = 15$ . Also, $M$ is the midpoint of side $AB$ and $H$ is the foot of the altitude from $A$ to $BC$ .
The length of $HM$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, A=(0,6), B=(-4,0), C=(5,0), M=B+3.6*dir(B--A); draw(B--C--A--B^^M--H--A^^rightanglemark(A,H,C)); label("A", A, NE); label("B", B, W); label("C", C, E); label("H", H, S); label("M", M, dir(M)); [/asy]
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 6.5\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 7.5\qquad \textbf{(E)}\ 8$

Solution:
Warning: this solution is very intensive in calculation. Please do NOT try this on the test!
Let's start by finding $AH$ . By Heron's Formula, $s=\frac{13+14+15}{2}=21, [ABC]=\sqrt{21*(21-13)(21-14)(21-15)}=84$ . Using the area formula $A=0.5bh$ $AH=12$ . Now using the Pythagorean Theorem, $BH=5, HC=9$
Now $AM=MB=6.5$ . Using Stewart's Theorem on $\triangle{ABH}$ , letting $HM=x$
$13*6.5*6.5+13x^2=12*6.5*12+5*6.5*5$ (remember that Stewart's Theorem is $man+dad=bmb+cnc$ ).
Thus $x=6.5$ or $x=-6.5$ (reject this solution since $x$ is positive). Thus $HM=6.5$ . Select $\boxed{6.5}$


Answer: 289.0

In $\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overline{BM}$ at $E.$ The ratio $DE: EF$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution:
By the Angle Bisector Theorem, we know that $[CBD]=\frac{169}{289}[ABC]$ . Therefore, by finding the area of triangle $CBD$ , we see that \[\frac{507\cdot BD}{2}\sin\frac{B}{2}=\frac{169}{289}[ABC].\] Solving for $BD$ yields \[BD=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}}.\] Furthermore, $\cos\frac{B}{2}=\frac{BD}{BF}$ , so \[BF=\frac{BD}{\cos\frac{B}{2}}=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}\cos\frac{B}{2}}.\] Now by the identity $2\sin\frac{B}{2}\cos\frac{B}{2}=\sin B$ , we get \[BF=\frac{4[ABC]}{3\cdot289\sin B}.\] But then $[ABC]=\frac{360\cdot 507}{2}\sin B$ , so $BF=\frac{240}{289}\cdot 507$ . Thus $BF:FC=240:49$
Now by the Angle Bisector Theorem, $CD=\frac{169}{289}\cdot 780$ , and we know that $MC=\frac{1}{2}\cdot 780$ so $DM:MC=\frac{169}{289}-\frac{1}{2}:\frac{1}{2}=49:289$
We can now use mass points on triangle CBD. Assign a mass of $240\cdot 49$ to point $C$ . Then $D$ must have mass $240\cdot 289$ and $B$ must have mass $49\cdot 49$ . This gives $F$ a mass of $240\cdot 49+49\cdot 49=289\cdot 49$ . Therefore, $DE:EF=\frac{289\cdot 49}{240\cdot 289}=\frac{49}{240}$ , giving us an answer of $\boxed{289}.$

Solution: Let $\angle{DBM}=\theta$ and $\angle{DBC}=\alpha$ . Then because $BM$ is a median we have $360\sin{(\alpha+\theta)}=507\sin{(\alpha-\theta)}$ . Now we know \[\sin{(\alpha+\theta)}=\sin{\alpha}\cos{\theta}+\sin{\theta}\cos{\alpha}=\dfrac{DF\cdot BD}{BF\cdot BE}+\dfrac{DE\cdot BD}{BE\cdot BF}=\dfrac{BD(DF+DE)}{BF\cdot BE}\] Expressing the area of $\triangle{BEF}$ in two ways we have \[\dfrac{1}{2}BE\cdot BF\sin{(\alpha-\theta)}=\dfrac{1}{2}EF\cdot BD\] so \[\sin{(\alpha-\theta)}=\dfrac{EF\cdot BD}{BF\cdot BE}\] Plugging this in we have \[\dfrac{360\cdot BD(DF+DE)}{BF\cdot BE}=\dfrac{507\cdot BD\cdot EF}{BF\cdot BE}\] so $\dfrac{DF+DE}{EF}=\dfrac{507}{360}$ . But $DF=DE+EF$ , so this simplifies to $1+\dfrac{2DE}{EF}=\dfrac{507}{360}=\dfrac{169}{120}$ , and thus $\dfrac{DE}{EF}=\dfrac{49}{240}$ , and $m+n=\boxed{289}$

Solution:
Firstly, angle bisector theorem yields $\frac{CD}{AD} = \frac{507}{360} = \frac{169}{120}$ . We're given that $AM=MC$ . Therefore, the cross ratio
\[(A,C;M,D) = \frac{AM(CD)}{AD(MC)} = \frac{169}{120}\]
We need a fourth point for this cross ratio to be useful, so reflect point $F$ over angle bisector $BD$ to a point $F'$ . Then $\triangle BFF'$ is isosceles and $BD$ is an altitude so $DF = DF'$ . Therefore,
\[(A,C;M,D) = (F,F';D,E) \implies \frac{FD(EF')}{EF(DF')} = \frac{EF'}{EF} = \frac{169}{120}\]
All that's left is to fiddle around with the ratios:
\[\frac{EF'}{EF} = \frac{ED+DF'}{EF} = \frac{EF+2DF}{EF} = 1\ +\ 2\left(\frac{DF}{EF}\right) \implies \frac{DF}{EF} = \frac{49}{240} \implies \boxed{289}\]

Solution: Extend $DF$ to intersect with the extension of $AB$ at $G$ . Notice that $\triangle{BDF} \cong \triangle{BDG}$ , so $GD=DF$ . We now use Menelaus on $\triangle{GBF}$ , as $A$ $D$ , and $C$ are collinear; this gives us $\frac{GA}{BA} \cdot \frac{BC}{FC} \cdot \frac{DF}{GD}=1$ . As $GD=DF$ , we have $\frac{GA}{AB}=\frac{FC}{BC}$ , hence $\frac{GA}{120}=\frac{FC}{169}$ . Reflect $G$ over $A$ to $G'$ . Note that $\frac{G'A}{BA}=\frac{FC}{BC}$ , and reflexivity, hence $\triangle{ABC} \sim \triangle{BG'F}$ . It's easily concluded from this that $G'F \parallel AC$ , hence $G'F \parallel AD$ . As $GD=DF$ , we have $AD$ is a midsegment of $\triangle{GG'F}$ , thus $G'F = 2AD$ . We now focus on the ratio $\frac{BF}{BC}$ . From similarity, we have $\frac{BF}{BC}=\frac{G'F}{AC}=\frac{2AD}{AC}$ . By the angle bisector theorem, we have $AD:DC=120:169$ , hence $AD:AC=120:289$ , so $\frac{BF}{BC}=\frac{240}{289}$ . We now work out the ratio $\frac{DM}{MC}$ $\frac{DM}{MC}=\frac{CD-MC}{MC}=\frac{CD}{MC}-1=\frac{2CD}{AC}-1=\frac{338}{289}-1=\frac{49}{289}$ . We now use mass points on $\triangle{BDC}$ . We let the mass of $C$ be $240\cdot 49$ , so the mass of $B$ is $49 \cdot 49$ and the mass of $D$ is $289\cdot 240$ . Hence, the mass of $F$ is $289\cdot 49$ , so the ratio $\frac{DE}{EF}=\frac{49}{240}$ . Extracting gives $49+240=\boxed{289}.$


Answer: 9.0

In $\triangle ABC, AB = 8, BC = 7, CA = 6$ and side $BC$ is extended, as shown in the figure, to a point $P$ so that $\triangle PAB$ is similar to $\triangle PCA$ . The length of $PC$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, P=(1.5,5), B=(8,0), C=P+2.5*dir(P--B); draw(A--P--C--A--B--C); label("A", A, W); label("B", B, E); label("C", C, NE); label("P", P, NW); label("6", 3*dir(A--C), SE); label("7", B+3*dir(B--C), NE); label("8", (4,0), S); [/asy]
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$

Solution:
Since we are given that $\triangle{PAB}\sim\triangle{PCA}$ , we have $\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}$
Solving for $PA$ in $\frac{PC}{PA}=\frac{6}{8}=\frac{3}{4}$ gives us $PA=\frac{4PC}{3}$
We also have $\frac{PA}{PC+7}=\frac{3}{4}$ . Substituting $PA$ in for our expression yields $\frac{\frac{4PC}{3}}{PC+7}=\frac{3}{4}$
Which we can further simplify to $\frac{16PC}{3}=3PC+21$
$\frac{7PC}{3}=21$
$PC=9\implies\boxed{9}$


Answer: 289.0

In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution: By the Law of Cosines on $\triangle ABC$ , we have: \[\cos(A) = \frac{10^2+10^2-12^2}{2*10*10}  = \frac{7}{25}\] By the Law of Cosines on $\triangle ADE$ , then \[\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}\] So, our answer is $250+39=\boxed{289}$

Solution: We draw the altitude from $B$ to $\overline{AC}$ to get point $F$ . We notice that the triangle's height from $A$ to $\overline{BC}$ is 8 because it is a $3-4-5$ Right Triangle. To find the length of $\overline{BF}$ , we let $h$ represent $\overline{BF}$ and set up an equation by finding two ways to express the area. The equation is $(8)(12)=(10)(h)$ , which leaves us with $h=9.6$ . We then solve for the length $\overline{AF}$ , which is done through pythagorean theorm and get $\overline{AF}$ $2.8$ . We can now see that $\triangle AFB$ is a $7-24-25$ Right Triangle. Thus, we set $\overline{AG}$ as $5-$ $\tfrac{x}{2}$ , and yield that $\overline{AD}$ $=$ $($ $5-$ $\tfrac{x}{2}$ $)$ $($ $\tfrac{25}{7}$ $)$ . Now, we can see $x$ $($ $5-$ $\tfrac{x}{2}$ $)$ $($ $\tfrac{25}{7}$ $)$ . Solving this equation, we yield $39x=250$ , or $x=$ $\tfrac{250}{39}$ . Thus, our final answer is $250+39=\boxed{289}$

Solution:
We start by adding a few points to the diagram. Call $F$ the midpoint of $AE$ , and $G$ the midpoint of $BC$ . (Note that $DF$ and $AG$ are altitudes of their respective triangles). We also call $\angle BAC = \theta$ . Since triangle $ADE$ is isosceles, $\angle AED = \theta$ , and $\angle ADF = \angle EDF = 90 - \theta$ . Since $\angle DEA = \theta$ $\angle DEC = 180 - \theta$ and $\angle EDC = \angle ECD = \frac{\theta}{2}$ . Since $FDC$ is a right triangle, $\angle FDC = 180 - 90 - \frac{\theta}{2} = \frac{180-m}{2}$
Since $\angle BAG = \frac{\theta}{2}$ and $\angle ABG = \frac{180-m}{2}$ , triangles $ABG$ and $CDF$ are similar by Angle-Angle similarity. Using similar triangle ratios, we have $\frac{AG}{BG} = \frac{CF}{DF}$ $AG = 8$ and $BG = 6$ because there are $2$ $6-8-10$ triangles in the problem. Call $AD = x$ . Then $CE = x$ $AE = 10-x$ , and $EF = \frac{10-x}{2}$ . Thus $CF = x + \frac{10-x}{2}$ . Our ratio now becomes $\frac{8}{6} = \frac{x+ \frac{10-x}{2}}{DF}$ . Solving for $DF$ gives us $DF = \frac{30+3x}{8}$ . Since $DF$ is a height of the triangle $ADE$ $FE^2 + DF^2 = x^2$ , or $DF = \sqrt{x^2 - (\frac{10-x}{2})^2}$ . Solving the equation $\frac{30+3x}{8} = \sqrt{x^2 - (\frac{10-x}{2})^2}$ gives us $x = \frac{250}{39}$ , so our answer is $250+39 = \boxed{289}$

Solution: As in the diagram, let $DE = x$ . Consider point $G$ on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on $DG, GC$ , and $DC$ . Let $GE = \frac{10-x}{2}$ . Therefore, it is trivial to see that $GC^2 = \Big(x + \frac{10-x}{2}\Big)^2$ (leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle $DGE$ , we know that $DG^2 = x^2 - \Big(\frac{10-x}{2}\Big)^2$ . Finally, we apply Law of Cosines on Triangle $DBC$ . We know that $\cos(\angle DBC) = \frac{3}{5}$ . Therefore, we get that $DC^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}$ . We can now do our final calculation: \[DG^2 + GC^2 = DC^2 \implies x^2 - \Big(\frac{10-x}{2}\Big)^2 + \Big(x + \frac{10-x}{2}\Big)^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}\] After some quick cleaning up, we get \[30x = \frac{72}{5} + 100 \implies x = \frac{250}{39}\] Therefore, our answer is $250+39=\boxed{289}$

Solution: Let $B = (0, 0)$ $C = (12, 0)$ , and $A = (6, 8)$ . Then, let $x$ be in the interval $0<x<2$ and parametrically define $D$ and $E$ as $(6-3x, 8-4x)$ and $(12-3x, 4x)$ respectively. Note that $AD = 5x$ , so $DE = 5x$ . This means that \begin{align*} \sqrt{36+(8x-8)^2} &= 5x\\ 36+(8x-8)^2 &= 25x^2\\ 64x^2-128x+100 &= 25x^2\\ 39x^2-128x+100 &= 0\\ x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\ x &= \dfrac{100}{78}, 2\\ \end{align*} However, since $2$ is extraneous by definition, $x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}$ ~ mathwiz0803

Solution: As shown in the diagram, let $x$ denote $\overline{AD}$ . Let us denote the foot of the altitude of $A$ to $\overline{BC}$ as $F$ . Note that $\overline{AE}$ can be expressed as $10-x$ and $\triangle{ABF}$ is a $6-8-10$ triangle . Therefore, $\sin(\angle{BAF})=\frac{3}{5}$ and $\cos(\angle{BAF})=\frac{4}{5}$ . Before we can proceed with the Law of Cosines, we must determine $\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}$ . Using LOC, we can write the following statement: \[(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies\] \[x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies\] \[0=(10-x)^2-\frac{14x}{25}(10-x)\implies\] \[0=10-x-\frac{14x}{25}\implies\] \[10=\frac{39x}{25}\implies x=\frac{250}{39}\] Thus, the desired answer is $\boxed{289}$ ~ blitzkrieg21

Solution:
In isosceles triangle, draw the altitude from $D$ onto $\overline{AD}$ . Let the point of intersection be $X$ . Clearly, $AE=10-AD$ , and hence $AX=\frac{10-AD}{2}$
Now, we recognise that the perpendicular from $A$ onto $\overline{AD}$ gives us two $6$ $8$ $10$ triangles. So, we calculate $\sin \angle ABC=\frac{4}{5}$ and $\cos \angle ABC=\frac{3}{5}$
$\angle BAC = 180-2\cdot\angle ABC$ . And hence,
$\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC) = -\cos (2\cdot\angle ABC) = \sin^2 \angle ABC - \cos^2 \angle ABC = 2\cos^2 \angle ABC - 1 = \frac{32}{25}-\frac{25}{25}=\frac{7}{25}$
Inspecting $\triangle ADX$ gives us $\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}$ Solving the equation $\frac{10-x}{2x}=\frac{7}{25}$ gives $x= \frac{250}{39} \implies\boxed{289}$

Solution:
We can have 2 Law of Cosines applied on $\angle A$ (one from $\triangle ADE$ and one from $\triangle ABC$ ),
$x^2=x^2+(10-x)^2-2(x)(10-x)\cdot \cos{A}$ and $12^2=10^2+10^2-2(10)(10)\cdot \cos{A}$
Solving for $\cos{A}$ in both equations, we get
$\cos{A} = \frac{(10-x)^2}{(2)(10-x)(x)}$ and $cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies  x = \frac{250}{39}$ , so the answer is $\boxed{289}$

Solution:
Let $B=(0, 0)$ , and $C=(12, 0)$ . From there, we know that $A=(6, 8)$ , so line $AB$ is $y=\frac{4}{3}x$ . Hence, $D=(a, \frac{4}{3}a)$ for some $a$ , and $BD=\frac{5}{3}a$ so $AD=10-\frac{5}{3}a$ . Now, notice that by symmetry, $E=(6+a, 8-\frac{4}{3}a)$ , so $ED^2=6^2+(8-\frac{8}{3}a)^2$ . Because $AD=ED$ , we now have $(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2$ , which simplifies to $\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100$ , so $\frac{39}{9}a=\frac{13}{3}a=\frac{28}{3}$ , and $a=\frac{28}{13}$ .
It follows that $AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}$ , and our answer is $250+39=\boxed{289}$

Solution:
Let's label $\angle A = \theta$ and $\angle ECD = \alpha$ . Using isosceles triangle properties and the triangle angle sum equation, we get \[180-(180-2\theta+\alpha) + \frac{180-\theta}{2} + \left(\frac{180-\theta}{2} - \alpha\right) = 180.\] Solving, we find $\theta = 2 \alpha$
Relabelling our triangle, we get $\angle ABC = 90 - \alpha$ . Dropping an altitude from $A$ to $BC$ and using the Pythagorean theorem, we find $[ABC] = 48$ . Using the sine area formula, we see $\frac12 \cdot 10 \cdot 12 \cdot \sin(90-\alpha) = 48$ . Plugging in our sine angle cofunction identity, $\sin(90-\alpha) = \cos(\alpha)$ , we get $\alpha = \cos{^{-1}}{\frac45}$
Now, using the Law of Sines on $\triangle ADE$ , we get \[\frac{\sin{2\alpha}}{\frac{p}{q}} = \frac{\sin{(180-4\alpha)}}{10-\frac{p}{q}}.\] After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as $\sin{(180-4\alpha)}=\sin{4\alpha}$ and $\sin{\left(\cos{^{-1}}{\frac45}\right)} = \frac35$ , we find $\frac{p}{q} = \frac{10\sin{2\alpha}}{\sin{4\alpha}+\sin{2\alpha}} = \frac{250}{39}$
Therefore, our answer is $250 + 39 = \boxed{289}$

Solution: We start by labelling a few angles (all of them in degrees). Let $\angle{BAC}=2\alpha = \angle{AED}, \angle{EDC}=\angle{ECD}=\alpha, \angle{DEC}=180-2\alpha, \angle{BDC}=3\alpha, \angle{DCB}=90-2\alpha, \angle{DBC}=90-\alpha$ . Also let $AD=a$ . By sine rule in $\triangle{ADE},$ we get $\frac{a}{\sin{2\alpha}}=\frac{10-a}{\sin{4\alpha}} \implies \cos{2\alpha}=\frac{5}{a}-\frac{1}{2}$ Using sine rule in $\triangle{ABC}$ , we get $\sin{\alpha}=\frac{3}{5}$ . Hence we get $\cos{2\alpha}=1-2\sin^2{\alpha}=1-\frac{18}{25}=\frac{7}{25}$ . Hence $\frac{5}{a}=\frac{1}{2}+\frac{7}{25}=\frac{39}{50} \implies a=\frac{250}{39}$ . Therefore, our answer is $\boxed{289}$


Answer: 14.0

In $\triangle BAC$ $\angle BAC=40^\circ$ $AB=10$ , and $AC=6$ .  Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively.  What is the minimum possible value of $BE+DE+CD$
$\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sqrt 3+9\qquad$

Solution: Let $C_1$ be the reflection of $C$ across $\overline{AB}$ , and let $C_2$ be the reflection of $C_1$ across $\overline{AC}$ .  Then it is well-known that the quantity $BE+DE+CD$ is minimized when it is equal to $C_2B$ .  (Proving this is a simple application of the triangle inequality; for an example of a simpler case, see Heron's Shortest Path Problem.)  As $A$ lies on both $AB$ and $AC$ , we have $C_2A=C_1A=CA=6$ .  Furthermore, $\angle CAC_1=2\angle CAB=80^\circ$ by the nature of the reflection, so $\angle C_2AB=\angle C_2AC+\angle CAB=80^\circ+40^\circ=120^\circ$ .  Therefore by the Law of Cosines \[BC_2^2=6^2+10^2-2\cdot 6\cdot 10\cos 120^\circ=196\implies BC_2=\boxed{14}.\]

Solution:
In $\triangle BAC$ , the three lines look like the Chinese character 又. Let $\triangle DEA$ $\triangle CDA$ , and $\triangle BEA$ have bases $DE$ $CD$ , and $BE$ respectively. Then, $\triangle DEA$ has the same side $DA$ as $\triangle CDA$ and the same side $EA$ as $\triangle BEA$ . Connect all three triangles with $\triangle DEA$ in the center and the two triangles sharing one of its sides. Then, $\pentagon BACDE$ is formed with $BE+DE+CD$ forming the base.
Intuitively, the pentagon's base is minimized when all three bottom sides are collinear. This is simply the original $\triangle BAC$ except that $\angle BAC =120^\circ$ . (In $\triangle DEA$ $\triangle CDA$ , and $\triangle BEA$ $\angle A = 40^\circ$ , and the three triangles connect at $A$ to form the pentagon). Thus, $m\angle BAC = 40 \cdot 3$ ).
$BC$ in this new triangle is then the minimum of $BE+DE+CD$ . Applying law of cosines, $BC=\sqrt{6^2+10^2-2(6)(10)\cos (120^\circ)}=\sqrt{196}=14 \implies \boxed{14}$

Solution:
(Diagram by shihan)
Reflect $C$ across $AB$ to $C'$ . Similarly, reflect $B$ across $AC$ to $B'$ . Clearly, $BE = B'E$ and $CD = C'D$ . Thus, the sum $BE + DE + CD = B'E + DE + C'D$ . This value is minimized when $B'$ $C'$ $D$ and $E$ are collinear. To finish, we use the law of cosines on the triangle $AB'C'$ $B'C' = \sqrt{6^2 + 10^2 - 2(6)(10)\cos 120^{\circ}} = \boxed{14}.$


Answer: 80.0

In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad  \textbf{(B) } 77 \qquad  \textbf{(C) } 78 \qquad  \textbf{(D) } 79 \qquad \textbf{(E) } 80$

Solution:
Let $P$ be the origin, and $PA$ lie on the $x$ -axis.
We can find $U=\left(\cos(36), \sin(36)\right)$ and $G=\left(10-\cos(56), \sin(56)\right)$
Then, we have $M=(5, 0)$ and $N$ is the midpoint of $U$ and $G$ , or $\left(\frac{10+\cos(36)-\cos(56)}{2}, \frac{\sin(36)+\sin(56)}{2}\right)$
Notice that the tangent of our desired points is the the absolute difference between the $y$ -coordinates of the two points divided by the absolute difference between the $x$ -coordinates of the two points.
This evaluates to \[\frac{\sin(36)+\sin(56)}{\cos(36)-\cos(56)}\] Now, using sum to product identities, we have this equal to \[\frac{2\sin(46)\cos(10)}{-2\sin(46)\sin({-10})}=\frac{\sin(80)}{\cos(80)}=\tan(80)\] so the answer is $\boxed{80}.$

Solution:
We will refer to the Diagram section. In this solution, all angle measures are in degrees.
We rotate $\triangle PUM$ by $180^\circ$ about $M$ to obtain $\triangle AU'M.$ Let $H$ be the intersection of $\overline{PA}$ and $\overline{GU'},$ as shown below.  Note that $\triangle GU'A$ is an isosceles triangle with $GA=U'A=1,$ so $\angle AGU'=\angle AU'G=\frac{180-\angle GAU'}{2}=44.$ In $\triangle GHA,$ it follows that $\angle GHA=180-\angle GAH-\angle AGH=80.$
Since $\frac{UM}{UU'}=\frac{UN}{UG}=\frac12,$ we conclude that $\triangle UMN\sim\triangle UU'G$ by SAS, from which $\angle UMN=\angle UU'G$ and $\angle UNM=\angle UGU'.$ By the Converse of the Corresponding Angles Postulate, we deduce that $\overline{MN}\parallel\overline{U'G}.$
Finally, we have $\angle NMA=\angle GHA=\boxed{80}$ by the Corresponding Angles Postulate.

Solution:
Link $PN$ , extend $PN$ to $Q$ so that $QN=PN$ . Then link $QG$ and $QA$
$\because M,N$ are the midpoints of $PA$ and $PQ,$ respectively
$\therefore MN$ is the midsegment of $\bigtriangleup PAQ$
$\therefore \angle QAP=\angle NMP$
Notice that $\bigtriangleup PUN\cong \bigtriangleup QGN$
As a result, $QG=AG=UP=1$ $\angle AQG=\angle QAG$ $\angle GQN=\angle NPU$
Also, $\angle GQN+\angle QPA=\angle QPU+\angle QPA=\angle UPA=36^{\circ}$
As a result, $2\angle QAG=180^{\circ}-56^{\circ}-36^{\circ}=88^{\circ}$
Therefore, $\angle QAP=\angle QAG+\angle TAP=56^{\circ}+44^{\circ}=100^{\circ}$
Since we are asked for the acute angle between the two lines, the answer to this problem is $\boxed{80}$

Solution:
Let the mid-point of $\overline{AT}$ be $B$ and the mid-point of $\overline{GT}$ be $C$ .
Since $BC=CG-BG$ and $CG=AB-\frac{1}{2}$ , we can conclude that $BC=\frac{1}{2}$ .
Similarly, we can conclude that $BM-CN=\frac{1}{2}$ . Construct $\overline{ND}\parallel\overline{BC}$ and intersects $\overline{BM}$ at $D$ , which gives $MD=DN=\frac{1}{2}$ .
Since $\angle{ABD}=\angle{BDN}$ $MD=DN$ , we can find the value of $\angle{DMN}$ , which is equal to $\frac{1}{2}\angle T=44^{\circ}$ . Since $\overline{BM}\parallel\overline{PT}$ , which means $\angle{DMN}+\angle{MNP}+\angle{P}=180^{\circ}$ , we can infer that $\angle{MNP}=100^{\circ}$ .
As we are required to give the acute angle formed, the final answer would be $80^{\circ}$ , which is $\boxed{80}$

Solution: Let the bisector of $\angle ATP$ intersect $PA$ at $X.$ We have $\angle ATX = \angle PTX = 44^{\circ},$ so $\angle TXA = 80^{\circ}.$ We claim that $MN$ is parallel to this angle bisector, meaning that the acute angle formed by $MN$ and $PA$ is $80^{\circ},$ meaning that the answer is $\boxed{80}$

Solution:
Note that $X$ , the midpoint of major arc $PA$ on $(PAT)$ is the Miquel Point of $PUAG$ (Because $PU = AG$ ). Then, since $1 = \frac{UN}{NG} = \frac{PM}{MA}$ , this spiral similarity carries $M$ to $N$ . Thus, we have $\triangle XMN \sim \triangle XAG$ , so $\angle XMN = \angle XAG$
But, we have $\angle XAG = \angle PAG = \angle PAX = 56 - \frac{180 - \angle PXA}{2} =56 - \frac{180 - \angle T}{2} = 56 - \frac{\angle A + \angle P}{2} = 56 - \frac{56+36}{2} = 56 - 46 = 10$ ; thus $\angle XMN = 10$
Then, as $X$ is the midpoint of the major arc, it lies on the perpendicular bisector of $PA$ , so $\angle XMA = 90$ . Since we want the acute angle, we have $\angle NMA = \angle XMA - \angle XMN = 90 - 10 = 80$ , so the answer is $\boxed{80}$

Solution: The argument of the average of any two unit vectors is average of the arguments of the two vectors. Thereby, the acute angle formed is \[\frac{36^\circ{} + 180^\circ{} - 56^\circ{}}{2} = \boxed{80}.\]


Answer: 120.0

In $\triangle PQR$ $PR=15$ $QR=20$ , and $PQ=25$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , and points $E$ and $F$ lie on $\overline{PR}$ , with $PA=QB=QC=RD=RE=PF=5$ . Find the area of hexagon $ABCDEF$

Solution: We know the area of the hexagon $ABCDEF$ to be $\triangle PQR- \triangle PAF- \triangle BCQ- \triangle RED$ . Since $PR^2+RQ^2=PQ^2$ , we know that $\triangle PRQ$ is a right triangle. Thus the area of $\triangle PQR$ is $150$ . Another way to compute the area is \[\frac12 \cdot PQ\cdot RQ \sin \angle PQR = \frac12 \cdot 500 \cdot \sin \angle PQR=150 \implies \sin \angle PQR = \frac35.\] Then the area of $\triangle BCQ = \frac12 \cdot BQ \cdot CQ \cdot \sin \angle PQR= \frac{25}{2}\cdot \frac{3}{5}=\frac{15}{2}$ . Preceding in a similar fashion for $\triangle PAF$ , the area of $\triangle PAF$ is $10$ . Since $\angle ERD = 90^{\circ}$ , the area of $\triangle RED=\frac{25}{2}$ . Thus our desired answer is $150-\frac{15}{2}-10-\frac{25}{2}=\boxed{120}$

Solution:
Let $R$ be the origin. Noticing that the triangle is a 3-4-5 right triangle, we can see that $A=(4,12), B=(16,3), C=(15,0), D=(5,0), E=(0,5)$ , and $F=(0,10)$ . Using the shoelace theorem, the area is $\boxed{120}$ .
Shoelace theorem:Suppose the polygon $P$ has vertices $(a_1, b_1)$ $(a_2, b_2)$ , ... , $(a_n, b_n)$ , listed in clockwise order. Then the area of $P$ is

Solution:
Note that $\triangle{PQR}$ has area $150$ and is a $3$ $4$ $5$ right triangle. Then, by similar triangles, the altitude from $B$ to $QC$ has length $3$ and the altitude from $A$ to $FP$ has length $4$ , so $[QBC]+[DRE]+[AFP]=\frac{15}{2}+\frac{25}{2}+\frac{20}{2}=30$ , meaning that $[ABCDEF]=\boxed{120}$ .
-Stormersyle

Solution: Knowing that $\triangle{PQR}$ has area $150$ and is a $3$ $4$ $5$ triangle, we can find the area of the smaller triangles $\triangle{DRE}$ $\triangle{APF}$ , and $\triangle{CQB}$ and subtract them from $\triangle{PQR}$ to obtain our answer. First off, we know $\triangle{DRE}$ has area $12.5$ since it is a right triangle. To the find the areas of $\triangle{APF}$ and $\triangle{CQB}$ , we can use Law of Cosines ( $c^2 = a^2 + b^2 - 2ab\cos C$ ) to find the lengths of $AF$ and $CB$ , respectively. Computing gives $AF = \sqrt{20}$ and $CB = \sqrt{10}$ . Now, using Heron's Formula, we find $\triangle{APF} = 10$ and $\triangle{CQB} = 7.5$ . Adding these and subtracting from $\triangle{PQR}$ , we get $150 - (10 + 7.5 + 12.5) = \boxed{120}$ -Starsher


Answer: 56.0

In $\triangle RED$ $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$ $RD=1$ . Let $M$ be the midpoint of segment $\overline{RD}$ . Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$ . Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$ . Then $AE=\frac{a-\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a+b+c$

Solution: Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$ , so $\overline{AP}\parallel\overline{EM}$ . Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$ , and $\overline{PM}\parallel\overline{CD}$ . Thus, $APME$ is a parallelogram and $AE = PM = \frac{CD}{2}$ . We can then use coordinates. Let $O$ be the foot of altitude $RO$ and set $O$ as the origin. Now we notice special right triangles! In particular, $DO = \frac{1}{2}$ and $EO = RO = \frac{\sqrt{3}}{2}$ , so $D\left(\frac{1}{2}, 0\right)$ $E\left(-\frac{\sqrt{3}}{2}, 0\right)$ , and $R\left(0, \frac{\sqrt{3}}{2}\right).$ $M =$ midpoint $(D, R) = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$ and the slope of $ME = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{1 + 2\sqrt{3}}$ , so the slope of $RC = -\frac{1 + 2\sqrt{3}}{\sqrt{3}}.$ Instead of finding the equation of the line, we use the definition of slope: for every $CO = x$ to the left, we go $\frac{x(1 + 2\sqrt{3})}{\sqrt{3}} = \frac{\sqrt{3}}{2}$ up. Thus, $x = \frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{3}{4\sqrt{3} + 2} = \frac{3(4\sqrt{3} - 2)}{44} = \frac{6\sqrt{3} - 3}{22}.$ $DC = \frac{1}{2} - x = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{14 - 6\sqrt{3}}{22}$ , and $AE = \frac{7 - \sqrt{27}}{22}$ , so the answer is $\boxed{056}$


Answer: 44.0

In $\triangle{ABC}$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle{AGE}$ is equilateral. Then $\cos(C)$ 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. What is $m+n+p?$
$\textbf{(A) }44 \qquad \textbf{(B) }48 \qquad \textbf{(C) }52 \qquad \textbf{(D) }56 \qquad \textbf{(E) }60$

Solution:
Let $AG = AE = GE = CE = 1$ . Since $G$ is the centroid, $DG = \frac12$ $BG = 2$
\[\angle BGD = \angle AGE = 60^{\circ}\]
By the Law of Cosine in $\triangle BGD$
\[BD^2 = BG^2 + DG^2 - 2 \cdot BG \cdot DG \cdot \cos \angle BGD\]
\[BD = \sqrt {2^2 + \left( \frac{1}{2} \right)^2 - 2 \cdot 2 \cdot \frac12 \cdot \cos \angle BGD} = \frac{\sqrt{13}}{2}, \quad CD = \frac{\sqrt{13}}{2}\]
By the Law of Cosine in $\triangle ACD$
\[AD^2 = AC^2 + CD^2 - 2 \cdot AC \cdot CD \cdot \cos \angle C\]
\[\cos \angle C = \frac{ AC^2 + CD^2 - AD^2 }{ 2 \cdot AC \cdot CD } = \frac{ 2^2 + \left( \frac{\sqrt{13}}{2} \right)^2 - \left( \frac{3}{2} \right)^2 }{ 2 \cdot 2 \cdot \frac{\sqrt{13}}{2} } = \frac{ 5 \sqrt{13} }{26}\]
\[5 + 13 + 26 = \boxed{44}\]

Solution: Using reference triangle $\triangle AGE$ , we can let \[A=(1,0,0),G=(0,1,0),E=(0,0,1),C=(-1,0,2),D=(-\tfrac{1}{2},\tfrac{3}{2},0),B=(0,3,-2).\] If we move $A,B,C$ each over by $(1,0,-2)$ , leaving $\angle C$ unchanged, we have \[A=(2,0,-2),B=(1,3,-4),C=(0,0,0).\] The angle $\theta$ between vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ satisfies \[\cos\theta=\frac{(2)(1)+(0)(3)+(-2)(-4)}{\sqrt{\left[2^{2}+0^{2}+(-2)^{2}\right]\left[1^{2}+3^{2}+(-4)^{2}\right]}}=\frac{10}{\sqrt{8\cdot 26}}=\frac{10}{4\sqrt{13}}=\frac{5\sqrt{13}}{26},\] giving the answer, $5+13+26=\boxed{44}$


Answer: 45.0

In $\triangle{ABC}$ with $AB = 12$ $BC = 13$ , and $AC = 15$ , let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii . Then $\frac{AM}{CM} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$

Solution:
Let $AM = x$ , then $CM = 15 - x$ . Also let $BM = d$ Clearly, $\frac {[ABM]}{[CBM]} = \frac {x}{15 - x}$ . We can also express each area by the rs formula. Then $\frac {[ABM]}{[CBM]} = \frac {p(ABM)}{p(CBM)} = \frac {12 + d + x}{28 + d - x}$ . Equating and cross-multiplying yields $25x + 2dx = 15d + 180$ or $d = \frac {25x - 180}{15 - 2x}.$ Note that for $d$ to be positive, we must have $7.2 < x < 7.5$
By Stewart's Theorem , we have $12^2(15 - x) + 13^2x = d^215 + 15x(15 - x)$ or $432 = 3d^2 + 40x - 3x^2.$ Brute forcing by plugging in our previous result for $d$ , we have $432 = \frac {3(25x - 180)^2}{(15 - 2x)^2} + 40x - 3x^2.$ Clearing the fraction and gathering like terms, we get $0 = 12x^4 - 340x^3 + 2928x^2 - 7920x.$
Aside: Since $x$ must be rational in order for our answer to be in the desired form, we can use the Rational Root Theorem to reveal that $6x$ is an integer because we can divide the polynomial by $2$ . The only such $x$ in the above-stated range is $\frac {22}3$
Legitimately solving that quartic, note that $x = 0$ and $x = 15$ should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get $0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6)(3x - 22).$ The only solution in the desired range is thus $\frac {22}3$ . Then $CM = \frac {23}3$ , and our desired ratio $\frac {AM}{CM} = \frac {22}{23}$ , giving us an answer of $\boxed{045}$

Solution:
Let $AM = 2x$ and $BM = 2y$ so $CM = 15 - 2x$ . Let the incenters of $\triangle ABM$ and $\triangle BCM$ be $I_1$ and $I_2$ respectively, and their equal inradii be $r$ . From $r = \sqrt {(s - a)(s - b)(s - c)/s}$ , we find that
\begin{align*}r^2 & = \frac {(x + y - 6)( - x + y + 6)(x - y + 6)}{x + y + 6} & (1) \\ & = \frac {( - x + y + 1)(x + y - 1)( - x - y + 14)}{ - x + y + 14}. & (2) \end{align*}
Let the incircle of $\triangle ABM$ meet $AM$ at $P$ and the incircle of $\triangle BCM$ meet $CM$ at $Q$ . Then note that $I_1 P Q I_2$ is a rectangle. Also, $\angle I_1 M I_2$ is right because $MI_1$ and $MI_2$ are the angle bisectors of $\angle AMB$ and $\angle CMB$ respectively and $\angle AMB + \angle CMB = 180^\circ$ . By properties of tangents to circles $MP = (MA + MB - AB)/2 = x + y - 6$ and $MQ = (MB + MC - BC)/2 = - x + y + 1$ . Now notice that the altitude of $M$ to $I_1 I_2$ is of length $r$ , so by similar triangles we find that $r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)$ (3). Equating (3) with (1) and (2) separately yields
\begin{align*} 2y^2 - 30 = 2xy + 5x - 7y \\  2y^2 - 70 = - 2xy - 5x + 7y, \end{align*}
and adding these we have
\[4y^2 - 100 = 0\implies y = 5\implies x = 11/3 \\ \implies AM/MC = (22/3)/(15 - 22/3) = 22/23 \implies \boxed{045}.\]

Solution:
Let the incircle of $ABM$ hit $AM$ $AB$ $BM$ at $X_{1},Y_{1},Z_{1}$ , and let the incircle of $CBM$ hit $MC$ $BC$ $BM$ at $X_{2},Y_{2},Z_{2}$ . Draw the incircle of $ABC$ , and let it be tangent to $AC$ at $X$ . Observe that we have a homothety centered at A sending the incircle of $ABM$ to that of $ABC$ , and one centered at $C$ taking the incircle of $BCM$ to that of $ABC$ . These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is $AX_{1}/AX=CX_{2}/CX$
By standard computations, $AX=\dfrac{AB+AC-BC}{2}=7$ and $CX=\dfrac{BC+AC-AB}{2}=8$ . Now, let $AX_{1}=7x$ and $CX_{2}=8x$ . We will now go around and chase lengths. Observe that $BY_{1}=BA-AY_{1}=BA-AX_{1}=12-7x$ . Then, $BZ_{1}=12-7x$ . We also have $CY_{2}=CX_{2}=8x$ , so $BY_{2}=13-8x$ and $BZ_{2}=13-8x$
Observe now that $X_{1}M+MX_{2}=AC-15x=15(1-x)$ . Also, $X_{1}M-MX_{2}=MZ_{1}-MZ_{2}=BZ_{2}-BZ_{1}=BY_{2}-BY_{1}=(1-x)$ . Solving, we get $X_{1}M=8-8x$ and $MX_{2}=7-7x$ (as a side note, note that $AX_{1}+MX_{2}=X_{1}M+X_{2}C$ , a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range).
Now, we get $BM=BZ_{2}+Z_{2}M=BZ_{2}+MX_{2}=20-15x$ . To finish, we will compute area ratios. $\dfrac{[ABM]}{[CBM]}=\dfrac{AM}{MC}=\dfrac{8-x}{7+x}$ . Also, since their inradii are equal, we get $\dfrac{[ABM]}{[CBM]}=\dfrac{40-16x}{40-14x}$ . Equating and cross multiplying yields the quadratic $3x^{2}-8x+4=0$ , so $x=2/3,2$ . However, observe that $AX_{1}+CX_{2}=15x<15$ , so we take $x=2/3$ . Our ratio is therefore $\dfrac{8-2/3}{7+2/3}=\dfrac{22}{23}$ , giving the answer $\boxed{045}$

Solution:
Let the common inradius equal r, $BM = x$ $AM = y$ $MC = z$
From the prespective of $\triangle{ABM}$ and $\triangle{BMC}$ we get:
$S_{ABM} = rs = r \cdot (\frac{12+x+y}{2})$ $S_{BMC} = rs = r \cdot (\frac{13+x+z}{2})$
Add two triangles up, we get $\triangle{ABC}$
$S_{ABC} =  S_{ABM} +  S_{BMC} = r \cdot \frac{25+2x+y+z}{2}$
Since $y + z = 15$ , we get:
By drawing an altitude from $I_1$ down to a point $H_1$ and from $I_2$ to $H_2$ , we can get:
$r \cdot cot(\frac{\angle A}{2}) =r \cdot A H_1 = r \cdot \frac{12+y-x}{2}$ and
$r \cdot cot(\frac{\angle C}{2}) = r \cdot H_2 C =  r \cdot \frac{13+z-x}{2}$
Adding these up, we get:
$r \cdot (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) = \frac{25+y+z-2x}{2} = \frac{40-2x}{2} = 20-x$
Now, we have 2 values equal to r, we can set them equal to each other:
If we let R denote the incircle of ABC, note:
AC = $(cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R = 15$ and
$S_{ABC} = \frac{12+13+15}{2} \cdot R = 20 \cdot R$
By cross multiplying the equation above, we get:
We can find out x:
Now, we can find ratio of y and z:
The answer is $\boxed{045}$

Solution:
Let $CM=x, AM=rx, BM=d$ $x+rx=15\Rightarrow x=\frac{15}{1+r}$
Similar to Solution 1, we have \[r=\frac{[AMB]}{[CMB]}=\frac{12+rx+d}{13+x+d} \Rightarrow d=\frac{13r-12}{1-r}\] as well as \[12^2\cdot x + 13^2 rx=15x\cdot rx+15d^2 (\text{via Stewart's Theorem})\] \[\frac{(12^2 + 13^2r) \cdot 15}{1+r} - \frac{15r\cdot 15^2}{(1+r)^2}=\frac{15(13r-12)^2}{(1-r)^2}\] \[\frac{169r^2+88r+144}{(1+r)^2}=\frac{(13r-12)^2}{(1-r)^2}\] \[(169r^2+88r+144)((r^2+1)-2r)=(169r^2-312r+144)((r^2+1)+2r)\] \[(r^2+1)(400r)=2r(338r^2-224r+288)\] \[100(r^2+1)=169r^2-112r+144 \Rightarrow 69r^2-112r+44=(23r-22)(3r-2)=0\]
Since $d=\frac{13r-12}{1-r}>0$ , we have $r=\frac{22}{23} \longrightarrow \boxed{045}$

Solution: Let $BM = d, AM = x, CM = 15 - x$ . Observe that we have the equation by the incircle formula: \[\frac{[ABM]}{12 + AM + MB} = \frac{[CBM]}{13 + CM + MB} \implies \frac{AM}{CM} = \frac{12 + MB}{13 + MB} \implies \frac{x}{15 - x} = \frac{12 + d}{13 + d}.\] Now let $X$ be the point of tangency between the incircle of $\triangle ABC$ and $AC$ . Additionally, let $P$ and $Q$ be the points of tangency between the incircles of $\triangle ABM$ and $\triangle CBM$ with $AC$ respectively. Some easy calculation yields $AX = 7, CX = 8$ . By homothety we have \[\frac{AP}{7} = \frac{CQ}{8} \implies 8(AP) = 7(CQ) \implies 8(12 + x - d) = 7(13 + 15 - x - d) \implies d = 15x - 100.\] Substituting into the first equation derived earlier it is left to solve \[\frac{x}{15 - x} = \frac{15x - 88}{15x - 87} \implies 3x^2 - 40x + 132 \implies (x - 6)(3x - 22) = 0.\] Now $x = 6$ yields $d = -10$ which is invalid, hence $x = \frac{22}{3}$ so $\frac{AM}{CM} = \frac{\frac{22}{3}}{15 - \frac{22}{3}} = \frac{22}{23}.$ The requested sum is $22 + 23 = \boxed{45}$ . ~blueprimes


Answer: 72.0

In $\triangle{ABC}$ with side lengths $AB = 13$ $AC = 12$ , and $BC = 5$ , let $O$ and $I$ denote the circumcenter and incenter, respectively. A circle with center $M$ is tangent to the legs $AC$ and $BC$ and to the circumcircle of $\triangle{ABC}$ . What is the area of $\triangle{MOI}$
$\textbf{(A)}\ \frac52\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ \frac{13}{4}\qquad\textbf{(E)}\ \frac72$

Solution:
In this solution, let the brackets denote areas.
We place the diagram in the coordinate plane: Let $A=(12,0),B=(0,5),$ and $C=(0,0).$
Since $\triangle ABC$ is a right triangle with $\angle ACB=90^\circ,$ its circumcenter is the midpoint of $\overline{AB},$ from which $O=\left(6,\frac52\right).$ Note that the circumradius of $\triangle ABC$ is $\frac{13}{2}.$
Let $s$ denote the semiperimeter of $\triangle ABC.$ The inradius of $\triangle ABC$ is $\frac{[ABC]}{s}=\frac{30}{15}=2,$ from which $I=(2,2).$
Since $\odot M$ is also tangent to both coordinate axes, its center is at $M=(a,a)$ and its radius is $a$ for some positive number $a.$ Let $P$ be the point of tangency of $\odot O$ and $\odot M.$ As $\overline{OP}$ and $\overline{MP}$ are both perpendicular to the common tangent line at $P,$ we conclude that $O,M,$ and $P$ are collinear. It follows that $OM=OP-MP,$ or \[\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a.\] Solving this equation, we have $a=4,$ from which $M=(4,4).$
Finally, we apply the Shoelace Theorem to $\triangle MOI:$ \[[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{72}.\] Remark


Answer: 77.0

In $\triangle{ABC}, AB=10, \angle{A}=30^\circ$ , and $\angle{C=45^\circ}$ . Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\perp{BC}$ $\angle{BAD}=\angle{CAD}$ , and $BM=CM$ . Point $N$ is the midpoint of the segment $HM$ , and point $P$ is on ray $AD$ such that $PN\perp{BC}$ . Then $AP^2=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
Let us just drop the perpendicular from $B$ to $AC$ and label the point of intersection $O$ . We will use this point later in the problem.
As we can see,
$M$ is the midpoint of $BC$ and $N$ is the midpoint of $HM$
$AHC$ is a $45-45-90$ triangle, so $\angle{HAB}=15^\circ$
$AHD$ is $30-60-90$ triangle.
$AH$ and $PN$ are parallel lines so $PND$ is $30-60-90$ triangle also.
Then if we use those informations we get $AD=2HD$ and
$PD=2ND$ and $AP=AD-PD=2HD-2ND=2HN$ or $AP=2HN=HM$
Now we know  that $HM=AP$ , we can find for $HM$ which is simpler to find.
We can use point $B$ to split it up as $HM=HB+BM$
We can chase those lengths and we would get
$AB=10$ , so $OB=5$ , so $BC=5\sqrt{2}$ , so $BM=\dfrac{1}{2} \cdot BC=\dfrac{5\sqrt{2}}{2}$
We can also use Law of Sines:
\[\frac{BC}{AB}=\frac{\sin\angle A}{\sin\angle C}\] \[\frac{BC}{10}=\frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}}\implies BC=5\sqrt{2}\]
Then using right triangle $AHB$ , we have $HB=10 \sin 15^\circ$
So $HB=10 \sin 15^\circ=\dfrac{5(\sqrt{6}-\sqrt{2})}{2}$
And we know that $AP = HM = HB + BM = \frac{5(\sqrt6-\sqrt2)}{2} + \frac{5\sqrt2}{2} = \frac{5\sqrt6}{2}$
Finally if we calculate $(AP)^2$
$(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}$ . So our final answer is $75+2=77$
$m+n=\boxed{077}$

Solution:
Break our diagram into 2 special right triangle by dropping an altitude from $B$ to $AC$ we then get that \[AC=5+5\sqrt{3}, BC=5\sqrt{2}.\] Since $\triangle{HCA}$ is a 45-45-90,
\[HC=\frac{5\sqrt2+5\sqrt6}{2}\] $MC=\frac{BC}{2},$ \[HM=\frac{5\sqrt6}{2}\] \[HN=\frac{5\sqrt6}{4}\] We know that $\triangle{AHD}\simeq \triangle{PND}$ and are 30-60-90.
Thus, \[AP=2 \cdot HN=\frac{5\sqrt6}{2}.\]
$(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}$ . So our final answer is $75+2=\boxed{077}$

Solution: Draw the $45-45-90 \triangle AHC$ . Now, take the perpendicular bisector of $BC$ to intersect the circumcircle of $\triangle ABC$ and $AC$ at $F, L, G$ as shown, and denote $O$ to be the circumcenter of $\triangle ABC$ . It is not difficult to see by angle chasing that $AHBGO$ is cyclic, namely with diameter $AB$ . Then, by symmetry, $EH = HB$ and as $HB, OG$ are both subtended by equal arcs they are equal. Hence, $EH = GO$ . Now, draw line $HL$ and intersect it at $AC$ at point $K$ in the diagram. It is not hard to use angle chase to arrive at $AEOL$ a parallelogram, and from our length condition derived earlier, $AL \parallel HG$ . From here, it is clear that $AK = KG$ ; that is, $P$ is just the intersection of the perpendicular from $K$ down to $BC$ and $AD$ ! After this point, note that $AP = PF$ . It is easily derived that the circumradius of $\triangle ABC$ is $\frac{10}{\sqrt{2}}$ . Now, $APO$ is a $30-60-90$ triangle, and from here it is easy to arrive at the final answer of $\boxed{077}$ . ~awang11's sol

Solution:
Let $BO \perp AC, O \in AC.$
Let $ME \perp BC, E \in AD.$
$MB = MC, \angle C = 45^\circ \implies$ points $M, E, O$ are collinear.
$HN = NM, AH||NP||ME \implies AP = PE.$
In $\triangle ABO \hspace{10mm} \angle A = 30^\circ \implies AO = AB \cos 60^\circ  = 5 \sqrt{3}.$
In $\triangle AEO \hspace{10mm} \angle A = 15^\circ, \angle O = 90^\circ + 45^\circ = 135^\circ  \implies$ \[\angle AEO = 30^\circ \implies\] \[AE = AO \frac {\sin 135^\circ}{\sin 30^\circ}  = 5 \sqrt{3} \cdot \sqrt{2} = 5 \sqrt{6} \implies\] \[AP = 5 \sqrt {\frac {3}{2}} \implies AP^2 = \frac {75}{2} \implies  \boxed{077}.\] vladimir.shelomovskii@gmail.com, vvsss


Answer: 62.0

In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?
$\textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62$

Solution: Let the population of the town in $1991$ be $p^2$ . Let the population in $2001$ be $q^2+9$ . It follows that $p^2+150=q^2+9$ . Rearrange this equation to get $141=q^2-p^2=(q-p)(q+p)$ . Since $q$ and $p$ are both positive integers with $q>p$ $(q-p)$ and $(q+p)$ also must be, and thus, they are both factors of $141$ . We have two choices for pairs of factors of $141$ $1$ and $141$ , and $3$ and $47$ . Assuming the former pair, since $(q-p)$ must be less than $(q+p)$ $q-p=1$ and $q+p=141$ . Solve to get $p=70, q=71$ . Since $p^2+300$ is not a perfect square, this is not the correct pair. Solve for the other pair to get $p=22, q=25$ . This time, $p^2+300=22^2+300=784=28^2$ . This is the correct pair. Now, we find the percent increase from $22^2=484$ to $28^2=784$ . Since the increase is $300$ , the percent increase is $\frac{300}{484}\times100\%\approx\boxed{62}$

Solution: Since all the answer choices are around $50\%$ , we know the town's starting population must be around $600$ . We list perfect squares from $400$ to $1000$ \[441, 484, 529,576,625,676,729,784,841,900,961\] We see that $484$ and $784$ differ by $300$ , and we can confirm that $484$ is the correct starting number by noting that $484+150=634=25^2+9$ . Thus, the answer is $784/484-1\approx \boxed{62}$

Solution: Let the population of the town in 1991 be $a^2$ and the population in 2011 be $b^2$ . We know that $a^2+150+150=b^2 \implies a^2-b^2=-300 \implies b^2-a^2=300 \implies (b-a)(b+a)=300$ . Note that $b-a$ must be even. Testing, we see that $a=22$ and $b=28$ works, as $484+150-9=625=25^2$ , so $\frac{784-484}{484} \approx \boxed{62}$


Answer: 414.0

In January $1980$ the Mauna Loa Observatory recorded carbon dioxide $(CO2)$ levels of $338$ ppm (parts per million). Over the years the average $CO2$ reading has increased by about $1.515$ ppm each year. What is the expected $CO2$ level in ppm in January $2030$ ? Round your answer to the nearest integer.
$\textbf{(A)}\ 399\qquad \textbf{(B)}\ 414\qquad \textbf{(C)}\ 420\qquad \textbf{(D)}\ 444\qquad \textbf{(E)}\ 459$

Solution: This is a time period of $50$ years, so we can expect the ppm to increase by $50*1.515=75.75~76$ $76+338=\boxed{414}$

Solution: For each year that has passed, the ppm will increase by $1.515$ . In $2030$ , the CO2 would have increased by $50\cdot 1.515 \approx. 76,$ so the total ppm of CO2 will be $76 + 338 = \boxed{414}.$

Solution:
2030 - 1980 = 50 years.
338 + 50 * 1.515 = 338 + 75.75 = 413.75 for 2030 ppm level $=\boxed{414}$


Answer: 62.0

In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below.
\[\begin{array}{c@{\hspace{8em}} c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} c@{\hspace{6pt}}c@{\hspace{6pt}}c} \vspace{4pt} \text{Row 0: } &    &    &     &     &    &    & 1 &     &     &    &    &    &  \\\vspace{4pt} \text{Row 1: } &    &    &     &     &    & 1 &    & 1  &     &    &    &    &  \\\vspace{4pt} \text{Row 2: } &    &    &     &     & 1 &    & 2 &     & 1  &    &    &    &  \\\vspace{4pt} \text{Row 3: } &    &    &     &  1 &    & 3 &    & 3  &     & 1 &    &    &  \\\vspace{4pt} \text{Row 4: } &    &    & 1  &     & 4 &    & 6 &     & 4  &    & 1 &    &  \\\vspace{4pt} \text{Row 5: } &    & 1 &     & 5  &    &10&    &10 &     & 5 &    & 1 &  \\\vspace{4pt} \text{Row 6: } & 1 &    & 6  &     &15&    &20&     &15 &    & 6 &    & 1 \end{array}\] In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3 :4 :5$

Solution:
Consider what the ratio means. Since we know that they are consecutive terms, we can say \[\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.\]
Taking the first part, and using our expression for $n$ choose $k$ \[\frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}\] \[\frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!}\] \[\frac{1}{3(n-k+1)} = \frac{1}{4k}\] \[n-k+1 = \frac{4k}{3}\] \[n = \frac{7k}{3} - 1\] \[\frac{3(n+1)}{7} = k\] Then, we can use the second part of the equation. \[\frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!}\] \[\frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!}\] \[\frac{1}{4(n-k)} = \frac{1}{5(k+1)}\] \[\frac{4(n-k)}{5} = k+1\] \[\frac{4n}{5}-\frac{4k}{5} = k+1\] \[\frac{4n}{5} = \frac{9k}{5} +1.\] Since we know $k = \frac{3(n+1)}{7}$ we can plug this in, giving us \[\frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1\] \[4n = 9\left(\frac{3(n+1)}{7}\right)+5\] \[7(4n - 5) = 27n+27\] \[28n - 35 = 27n+27\] \[n = 62\] We can also evaluate for $k$ , and find that $k = \frac{3(62+1)}{7} = 27.$ Since we want $n$ , however, our final answer is $\boxed{062.}$ $\LaTeX$ by ciceronii

Solution:
Call the row $x=t+k$ , and the position of the terms $t-1, t, t+1$ . Call the middle term in the ratio $N = \dbinom{t+k}{t} = \frac{(t+k)!}{k!t!}$ . The first term is $N \frac{t}{k+1}$ , and the final term is $N \frac{k}{t+1}$ . Because we have the ratio $3:4:5$
$\frac{t}{k+1} = \frac{3}{4}$ and $\frac{k}{t+1} = \frac{5}{4}$
$4t = 3k+3$ and $4k= 5t+5$
$4t-3k=3$ $5t-4k=-5$
Solve the equations to get $t= 27, k=35$ and $x = t+k = \boxed{062}$


Answer: 40.0

In Theresa's first $8$ basketball games, she scored $7, 4, 3, 6, 8, 3, 1$ and $5$ points. In her ninth game, she scored fewer than $10$ points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than $10$ points and her points-per-game average for the $10$ games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?
$\textbf{(A)}\ 35\qquad \textbf{(B)}\ 40\qquad \textbf{(C)}\ 48\qquad \textbf{(D)}\ 56\qquad \textbf{(E)}\ 72$

Solution: The total number of points from the first $8$ games is $7+4+3+6+8+3+1+5=37$ . We have to make this a multiple of $9$ by scoring less than $10$ points. The closest multiple of $9$ is $45$ $45-37=8$ . Now we have to add a number to get a multiple of 10. The next multiple is $50$ we added $5$ , multiplying these together you get $8\cdot5$ is $40$ . The answer is $\boxed{40}$


Answer: 860.0

In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column.  Let $N$ be the number of shadings with this property.  Find the remainder when $N$ is divided by 1000.
AIME I 2007-10.png

Solution:
Consider the first column.  There are ${6\choose3} = 20$ ways that the rows could be chosen, but without loss of generality let them be the first three rows.  (Change the order of the rows to make this true.)  We will multiply whatever answer we get by 20 to get our final answer.
Now consider the 3x3 that is next to the 3 boxes we have filled in.  We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each).  We split into three cases:
So there are $20(3+54+36) = 1860$ different shadings, and the solution is $\boxed{860}$

Solution:
We start by showing that every group of $6$ rows can be grouped into $3$ complementary pairs.
We proceed with proof by contradiction. Without loss of generality, assume that the first row has columns $1$ and $2$ shaded. Note how if there is no complement to this, then all the other five rows must have at least one square in the first two columns shaded. That means that in total, the first two rows have $2+5=7$ squares shaded in- that is false since it should only be $6$ .
Thus, there exists another row that is complementary to the first. We remove those two and use a similar argument again to show that every group of $6$ rows can be grouped into $3$ complementary pairs.
Now we proceed with three cases.
Our answer is thus $720+1080+60=1860,$ leaving us with a final answer of $\boxed{860}.$

Solution:
Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.
There are ${6 \choose 3}$ ways to choose which rows have 1 shaded square (which we'll call a "1-row") within the first 3 columns and which rows have 2 (we'll call these "2-rows") within the first 3 columns. Next, we do some casework:
In total, we have ${6\choose3}(1+18+54+20)=20*93=1860$ . Thus our answer is $\boxed{860}$

Solution:
We can use generating functions. Suppose that the variables $a$ $b$ $c$ , and $d$ represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function $ab+ac+ad+bc+bd+cd$ , which we can write as $P(a,b,c,d)=(ab+cd)+(a+b)(c+d)$ . Therefore, $P(a,b,c,d)^6$ represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of $a^3b^3c^3d^3$ in $P(a,b,c,d)^6$
By the Binomial Theorem, \[P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}\] If we expand $(ab+cd)^k$ , then the powers of $a$ and $b$ are always equal. Therefore, to obtain terms of the form $a^3b^3c^3d^3$ , the powers of $a$ and $b$ in $(a+b)^{6-k}$ must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that $k$ must be even. We can use the same logic for $c$ and $d$ . Therefore, the coefficient of $a^3b^3c^3d^3$ in the following expression is the same as the coefficient of $a^3b^3c^3d^3$ in (1). \[\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}\] Now we notice that the only way to obtain terms of the form $a^3b^3c^3d^3$ is if we take the central term in the binomial expansion of $(ab+cd)^{2k}$ . Therefore, the terms that contribute to the coefficient of $a^3b^3c^3d^3$ in (2) are \[\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.\] This sum is $400+1080+360+20=1860$ so the answer is $\boxed{860}$