Answer: 5.0

sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m + n.$

Solution:
The center $I$ of the insphere must be located at $(r,r,r)$ where $r$ is the sphere's radius. $I$ must also be a distance $r$ from the plane $ABC$
The signed distance between a plane and a point $I$ can be calculated as $\frac{(I-G) \cdot P}{|P|}$ , where G is any point on the plane, and P is a vector perpendicular to ABC.
A vector $P$ perpendicular to plane $ABC$ can be found as $V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle$
Thus $\frac{(I-C) \cdot P}{|P|}=-r$ where the negative comes from the fact that we want $I$ to be in the opposite direction of $P$
\begin{align*}\frac{(I-C) \cdot P}{|P|}&=-r\\ \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&=-r\\ \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&=-r\\ \frac{44r -48}{28}&=-r\\ 44r-48&=-28r\\ 72r&=48\\ r&=\frac{2}{3} \end{align*}
Finally $2+3=\boxed{005}$

Solution: Notice that we can split the tetrahedron into $4$ smaller tetrahedrons such that the height of each tetrahedron is $r$ and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the tetrahedrons are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be $V$ and surface area be $F$ , using the volume formula for each pyramid(base times height divided by 3) we have $\dfrac{rF}{3}=V$ . The surface area of the pyramid is $\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]$ . We know triangle ABC's side lengths, $\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},$ and $\sqrt{4^{2}+6^{2}}$ , so using the expanded form of heron's formula, \begin{align*}[ABC]&=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}\\ &=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}\\ &=\sqrt{196}\\ &=14\end{align*} Therefore, the surface area is $14+22=36$ , and the volume is $\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8$ , and using the formula above that $\dfrac{rF}{3}=V$ , we have $12r=8$ and thus $r=\dfrac{2}{3}$ , so the desired answer is $2+3=\boxed{005}$

Solution: First let us find the equation of the plane passing through $(6,0,0), (0,0,2), (0,4,0)$ .  The "point-slope form" is $A(6-x1)+B(0-y1)+C(0-z1)=0.$ Plugging in $(0,0,2)$ gives $A(6)+B(0)+C(-2)=0.$ Plugging in $(0,4,0)$ gives $A(6)+B(-4)+C(0)=0.$ We can then use Cramer's rule/cross multiplication to get $A/(0-8)=-B/(0+12)=C/(-24)=k.$ Solve for A, B, C to get $2k, 3k, 6k$ respectively.  We can then get $2k(x-x1)+3k(y-y1)+6k(z-z1)=0.$ Cancel out k on both sides. Next, let us substitute $(0,0,2)$ .  We can then get $2x+3y+6z=12$ as the equation of the plane.  We can divide the equation by its magnitude to get the normal form of the plane.  We get $2x/7+2y/7+6z/7=12/7$ to be the normal form.  Note that the point is going to be at $(r,r,r).$ We find the distance from $(r,r,r)$ to the plane as $2/7r+3/7r+6/7r-12/7/(\sqrt{(4/49+9/49+36/49)})$ , which is $+/-(11r/7-12/7)$ .  We take the negative value of this because if we plug in $(0,0,0)$ to the equation of the plane we get a negative value.  We equate that value to r and we get the equation $-(11r/7-12/7)=r$ to solve $r={2/3}$ , so the answer is $\boxed{005}$

Solution:
Clearly, if the radius of the sphere is $r$ , the center of the sphere lies on $(r, r, r)$
We find the equation of plane $ABC$ to be $\frac16 x+\frac14 y+\frac12 z=1$ . From the definition of the insphere, it must be true that the distance from the center of the sphere to plane $ABC$ is equal to the length of the radius of the sphere. By point-to-plane, we have \[r=\frac{|\frac16 r+\frac14 r+\frac12 r-1|}{\sqrt{\left(\frac16\right)^2+\left(\frac14\right)^2+\left(\frac12\right)^2}} \implies r=\frac23,\] so the answer is $\boxed{005}$


Answer: 62.0

subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$
$\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68$

Solution: The question asks for the maximum possible number of elements. The integers from $1$ to $24$ can be included because you cannot make $125$ with integers from $1$ to $24$ without the other number being greater than $100$ . The integers from $25$ to $100$ are left. They can be paired so the sum is $125$ $25+100$ $26+99$ $27+98$ $\ldots$ $62+63$ . That is $38$ pairs, and at most one number from each pair can be included in the set. The total is $24 + 38 = \boxed{62}$

Solution: "Cut" $125$ into half. The maximum integer value in the smaller half is $62$ . Thus the answer is $\boxed{62}$

Solution: The maximum possible number of elements includes the smallest numbers. So, subset $B = \{1,2,3....n-1,n\}$ where n is the maximum number of elements in subset $B$ . So, we have to find two consecutive numbers, $n$ and $n+1$ , whose sum is $125$ . Setting up our equation, we have $n+(n+1) = 2n+1 = 125$ . When we solve for $n$ , we get $n =\boxed{62}$

Solution: We can put all odd numbers into subset B, or we can put all even numbers into subset B, so now there are 50 numbers in the set. I will use all even numbers in this solution. Now, we need to add other odd(or even!) numbers possible in this subset, which is all odd(or even) numbers that can be added so that the sum with 100(or 99) plus the biggest possible odd number(or even) to get 123. This will get us the numbers 1,3,5...21,23(or numbers 2,4,6...22,24), which gives us 12 more numbers, and adding that to the 50 original numbers, we get $B =\boxed{62}$


Answer: 89.0

triangle has vertices $P_{}^{}=(-8,5)$ $Q_{}^{}=(-15,-19)$ , and $R_{}^{}=(1,-7)$ . The equation of the bisector of $\angle P$ can be written in the form $ax+2y+c=0_{}^{}$ . Find $a+c_{}^{}$

Solution:
Use the angle bisector theorem to find that the angle bisector of $\angle P$ divides $QR$ into segments of length $\frac{25}{x} = \frac{15}{20 -x} \Longrightarrow x = \frac{25}{2},\ \frac{15}{2}$ . It follows that $\frac{QP'}{RP'} = \frac{5}{3}$ , and so $P' = \left(\frac{5x_R + 3x_Q}{8},\frac{5y_R + 3y_Q}{8}\right) = (-5,-23/2)$
The desired answer is the equation of the line $PP'$ $PP'$ has slope $\frac{-11}{2}$ , from which we find the equation to be $11x + 2y + 78 = 0$ . Therefore, $a+c = \boxed{089}$

Solution:
Transform triangle $PQR$ so that $P$ is at the origin. Note that the slopes do not change when we transform the triangle.
We know that the slope of $PQ$ is $\frac{24}{7}$ and the slope of $PR$ is $-\frac{4}{3}$ . Thus, in the complex plane, they are equivalent to $\tan(\alpha)=\frac{24}{7}$ and $\tan(\beta)=-\frac{4}{3}$ , respectively. Here $\alpha$ is the angle formed by the $x$ -axis and $PQ$ , and $\beta$ is the angle formed by the $x$ -axis and $PR$ . The equation of the angle bisector is $\tan\left(\frac{\alpha+\beta}{2}\right)$
As the tangents are in very neat Pythagorean triples , we can easily calculate $\cos(\alpha)$ and $\cos(\beta)$
Angle $\alpha$ is in the third quadrant, so $\cos(\alpha)$ is negative. Thus $\cos(\alpha)=-\frac{7}{25}$
Angle $\beta$ is in the fourth quadrant, so $\cos(\beta)$ is positive. Thus $\cos(\beta)=\frac{3}{5}$
By the Half-Angle Identities $\tan\left(\frac{\alpha}{2}\right)=\pm\sqrt{\frac{1-\cos(\alpha)}{1+\cos(\alpha)}}=\pm\sqrt{\frac{\frac{32}{25}}{\frac{18}{25}}}=\pm\sqrt{\frac{16}{9}}=\pm\frac{4}{3}$ and $\tan\left(\frac{\beta}{2}\right)=\pm\sqrt{\frac{1-\cos(\beta)}{1+\cos(\beta)}}=\pm\sqrt{\frac{\frac{2}{5}}{\frac{8}{5}}}=\pm\sqrt{\frac{1}{4}}=\pm\frac{1}{2}$
Since $\frac{\alpha}{2}$ and $\frac{\beta}{2}$ must be in the second quadrant, their tangent values are both negative. Thus $\tan\left(\frac{\alpha}{2}\right)=-\frac{4}{3}$ and $\tan\left(\frac{\beta}{2}\right)=-\frac{1}{2}$
By the sum of tangents formula $\tan\left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=\frac{\tan\left(\frac{\alpha}{2}\right)+\tan\left(\frac{\beta}{2}\right)}{1-\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)}=\frac{-\frac{11}{6}}{\frac{1}{3}}=-\frac{11}{2}$ , which is the slope of the angle bisector.
Finally, the equation of the angle bisector is $y-5=-\frac{11}{2}(x+8)$ or $y=-\frac{11}{2}x-39$ . Rearranging, we get $11x+2y+78=0$ , so our sum is $a+c=11+78=\boxed{089}$ . ~eevee9406


Answer: 17.0

triangular array of numbers has a first row consisting of the odd integers $1,3,5,\ldots,99$ in increasing order.  Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry.  Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it.  How many entries in the array are multiples of $67$

Solution:
Let the $k$ th number in the $n$ th row be $a(n,k)$ . Writing out some numbers, we find that $a(n,k) = 2^{n-1}(n+2k-2)$
We wish to find all $(n,k)$ such that $67| a(n,k) = 2^{n-1} (n+2k-2)$ . Since $2^{n-1}$ and $67$ are relatively prime , it follows that $67|n+2k-2$ . Since every row has one less element than the previous row, $1 \le k \le 51-n$ (the first row has $50$ elements, the second $49$ , and so forth; so $k$ can range from $1$ to $50$ in the first row, and so forth). Hence
it follows that $67| n - 2k + 2$ implies that $n-2k+2 = 67$ itself.
Now, note that we need $n$ to be odd, and also that $n+2k-2 = 67 \le 100-n \Longrightarrow n \le 33$
We can check that all rows with odd $n$ satisfying $1 \le n \le 33$ indeed contains one entry that is a multiple of $67$ , and so the answer is $\frac{33+1}{2} = \boxed{017}$

Solution: The result above is fairly intuitive if we write out several rows and then divide all numbers in row $r$ by $2^{r-1}$ (we can do this because dividing by a power of 2 doesn't affect divisibility by $67$ ). The second row will be $2, 4, 6, \cdots , 98$ , the third row will be $3, 5, \cdots, 97$ , and so forth. Clearly, only the odd-numbered rows can have a term divisible by $67$ . However, with each row the row will have one less element, and the $99-67+1 = 33$ rd row is the last time $67$ will appear. Therefore the number of multiples of 67 in the entire array is $\frac{33+1}{2} = \boxed{017}$


Answer: 640.0

triangular array of squares has one square in the first row, two in the second, and in general, $k$ squares in the $k$ th row for $1 \leq k \leq 11.$ With the exception of the bottom row, each square rests on two squares in the row immediately below (illustrated in the given diagram). In each square of the eleventh row, a $0$ or a $1$ is placed. Numbers are then placed into the other squares, with the entry for each square being the sum of the entries in the two squares below it. For how many initial distributions of $0$ 's and $1$ 's in the bottom row is the number in the top square a multiple of $3$
[asy] for (int i=0; i<12; ++i){  for (int j=0; j<i; ++j){    //dot((-j+i/2,-i));    draw((-j+i/2,-i)--(-j+i/2+1,-i)--(-j+i/2+1,-i+1)--(-j+i/2,-i+1)--cycle);  } } [/asy]

Solution:
Label each of the bottom squares as $x_0, x_1 \ldots x_9, x_{10}$
Through induction , we can find that the top square is equal to ${10\choose0}x_0 + {10\choose1}x_1 + {10\choose2}x_2 + \ldots {10\choose10}x_{10}$ . (This also makes sense based on a combinatorial argument: the number of ways a number can "travel" to the top position going only up is equal to the number of times it will be counted in the final sum.)
Examine the equation $\mod 3$ . All of the coefficients from $x_2 \ldots x_8$ will be multiples of $3$ (since the numerator will have a $9$ ). Thus, the expression boils down to $x_0 + 10x_1 + 10x_9 + x_{10} \equiv 0 \mod 3$ . Reduce to find that $x_0 + x_1 + x_9 + x_{10} \equiv 0 \mod 3$ . Out of $x_0,\ x_1,\ x_9,\ x_{10}$ , either all are equal to $0$ , or three of them are equal to $1$ . This gives ${4\choose0} + {4\choose3} = 1 + 4 = 5$ possible combinations of numbers that work.
The seven terms from $x_2 \ldots x_8$ can assume either $0$ or $1$ , giving us $2^7$ possibilities. The answer is therefore $5 \cdot 2^7 = \boxed{640}$