Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution: Define a coordinate system with $D = (0,0,0)$ $M = (1, \frac{1}{2}, 0)$ $N = (0,1,\frac{1}{2})$ . The plane formed by $D$ $M$ , and $N$ is $z = \frac{y}{2} - \frac{x}{4}$ . It intersects the base of the unit cube at $y = \frac{x}{2}$ . The z-coordinate of the plane never exceeds the height of the unit cube for $0 \leq x \leq 1, 0 \leq y \leq 1$ . Therefore, the volume of one of the two regions formed by the plane is \[\int_0^1 \int_{\frac{x}{2}}^1 \int_0^{\frac{y}{2}-\frac{x}{4}}dz\,dy\,dx = \frac{7}{48}\] Since $\frac{7}{48} < \frac{1}{2}$ , our answer is $1-\frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089}$

Answer:  089
Problem: Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution: Same coordinate system, but we will use domains instead of using triple integrals. By the way, the method to obtain the equation of the plane is the cross product ( $\vec{[1,.5,0]}\times\vec{[0,1,.5]}=\vec{[.25,-.5,1]}$ ). We can multiply this vector by $-4$ to make things look cleaner and get $\vec{[-1,2,-4]}$ . We then get the desired plane, $-x+2y-4z=0$ , or $z=\frac{2y-x}{4}$ . We use a double integral with a Type I domain. Observing the diagram, the domain is where $0 \leq x \leq 1, .5x \leq y \leq 1$ . The integral is then \[.25\int_0^1\int_{.5x}^1 2y - x\] which becomes \[.25\int_0^1 (y^2-xy)|_{.5x}^1\] which becomes \[.25\int_0^1 1-x+.25x^2\] which then becomes \[.25(1-\frac{1}{2}+\frac{1}{12})\] and finally \[\frac{7}{48}\] So our answer is $1^3-\frac{7}{48} = \frac{41}{48}, 41+48 = \boxed{089}$

Answer:  089
Problem: Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution: Let $Q$ be the intersection of the plane with edge $FB,$ then $\triangle MQB$ is similar to $\triangle DNC$ and the volume $[DNCMQB]$ is a sum of areas of cross sections of similar triangles running parallel to face $ABFE.$ Let $x$ be the distance from face $ABFE,$ let $h$ be the height parallel to $AB$ of the cross-sectional triangle at that distance, and $a$ be the area of the cross-sectional triangle at that distance. $A(x)=\frac{h^2}{4},$ and $h=\frac{x+1}{2},$ then $A=\frac{(x+1)^2}{16}$ , and the volume $[DNCMQB]$ is $\int^1_0{A(x)}\,\mathrm{d}x=\int^1_0{\frac{(x+1)^2}{16}}\,\mathrm{d}x=\frac{7}{48}.$ Thus the volume of the larger solid is $1-\frac{7}{48}=\frac{41}{48} \rightarrow p+q = \boxed{089}$

Answer:  089
Problem: Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution: The volume of a frustum is $\frac{h_2b_2 -h_1b_1}3$ where $b_i$ is the area of the base and $h_i$ is the height from the chopped off apex to the base.
We can easily see that from symmetry, the area of the smaller front base is $\frac{1}{16}$ and the area of the larger back base is $\frac{1}4$
Now to find the height of the apex.
Extend the $DM$ and (call the intersection of the plane with $FB$ G) $NG$ to meet at $x$ . Now from similar triangles $XMG$ and $XDN$ we can easily find the total height of the triangle $XDN$ to be $2$
Now straight from our formula, the volume is $\frac{7}{48}$ Thus the answer is:
$1-\text{Volume} \Longrightarrow \boxed{089}$

Answer:  089
Problem: Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution: We will solve for the area of the smaller region, and then subtract it from 1.
Let $X$ be the point where plane $DMN$ intersects $FB$ . Then $DMBCNX$ can be split into triangular pyramid $DMBX$ and quadrilateral pyramid $BCNXD$
Pyramid $DMBX$ has base $DMB$ with area $\frac{1}2 \cdot 1 \div 2 = \frac{1}4$ . The height is $BX = \frac{1}4$ , so the volume of $DMBX$ is $\frac{1}4 \cdot \frac{1}4 \div 3 = \frac{1}{48}$
Similarly, pyramid $BCNXD$ has base $BCNX$ with area $(\frac{1}4 + \frac{1}2) \cdot 1 = \frac{3}8$ . The height is $CD = 1$ , so the volume of $BCNXD$ is $\frac{3}8 \cdot 1 \div 3 = \frac{1}8$
Adding up the volumes of $DMBX$ and $BCNXD$ , we find that the volume of $DMBCNX$ is $\frac{1}{48} + \frac{6}{48} = \frac{7}{48}$ . Therefore the volume of the larger solid is $1 - \frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089}$

Answer:  089
Problem: Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution: We use a coordinate system with $C = (0,0,0)$ $D = (1,0,0)$ $M = (\frac{1}{2},1,0)$ $N = (0,0,\frac{1}{2})$ . Then the plane going through $D$ $M$ , and $N$ has equation $z = \frac{1}{2} - \frac{1}{2}x - \frac{1}{4}y$ . We set up a double integral in this coordinate system. Consider the region to be integrated over in the $xy$ -plane. From $x=0$ to $x=\frac{1}{2}$ , the upper bound of the region is $y = 1$ . From $x= \frac{1}{2}$ to $x=1$ , the upper bound of the region is $y = 2 - 2x$ . In both cases, the lower bound of the region is $y = 0$ . Thus, we have the double integral $\int_{0}^{\frac{1}{2}} \int_{0}^{1} \frac{1}{2} - \frac{1}{2}x - \frac{1}{4}y dydx + \int_{\frac{1}{2}}^{1} \int_{0}^{2-2x} \frac{1}{2} - \frac{1}{2}x - \frac{1}{4}y dydx$ . We find that this sum evaluates to $\frac{7}{48}$ . However, this is the volume of the smaller region, so the larger region's volume is that of the cube minus that of the smaller region. Since the cube has side length $1$ , its volume is $1$ , so the volume of the larger region is $1 - \frac{7}{48} = \frac{41}{48}$ . Thus, our answer is $41 + 48 = \boxed{089}$

Answer:  089
Problem: Let $x,$ $y,$ and $z$ be positive real numbers that satisfy \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.\] The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution: Since there are only two dependent equations given and three unknowns, the three expressions given can equate to any common value (that isn't 0, of course), so to simplify the problem let us assume without loss of generality that \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) = 2.\] Then \begin{align*} 2\log_{x}(2y) = 2 &\implies x=2y\\ 2\log_{2x}(4z) = 2 &\implies 2x=4z\\ \log_{2x^4}(8yz) = 2 &\implies 4x^8 = 8yz \end{align*} Solving these equations, we quickly see that $4x^8 = (2y)(4z) = x(2x) \rightarrow x=2^{-1/6}$ and then $y=z=2^{-1/6 - 1} = 2^{-7/6}.$ Finally, our desired value is $2^{-1/6} \cdot (2^{-7/6})^5 \cdot 2^{-7/6} = 2^{-43/6}$ and thus $p+q = 43 + 6 = \boxed{049.}$

Answer:  049.
Problem: Let $x,$ $y,$ and $z$ be positive real numbers that satisfy \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.\] The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution: Notice that $2y\cdot 4z=8yz$ $2\log_2(2y)=\log_2\left(4y^2\right)$ and $2\log_2(4z)=\log_2\left(16z^2\right)$
From this, we see that $8yz$ is the geometric mean of $4y^2$ and $16z^2$ . So, for constant $C\ne 0$ \[\frac{\log 4y^2}{\log x}=\frac{\log 8yz}{\log 2x^4}=\frac{\log 16z^2}{\log 2x}=C\] Since $\log 4y^2,\log 8yz,\log 16z^2$ are in an arithmetic progression, so are $\log x,\log 2x^4,\log 2x$
Therefore, $2x^4$ is the geometric mean of $x$ and $2x$ \[2x^4=\sqrt{x\cdot 2x}\implies 4x^8=2x^2\implies 2x^6=1\implies x=2^{-1/6}\] We can plug $x$ in to any of the two equal fractions aforementioned. So, without loss of generality: \[\frac{\log 4y^2}{\log x}=\frac{\log 16z^2}{\log 2x}\implies \log\left(4y^2\right)\log\left(2x\right)=\log\left(16z^2\right)\log\left(x\right)\] \[\implies \log\left(4y^2\right)\cdot \frac{5}{6}\log 2=\log\left(16z^2\right)\cdot \frac{-1}{6}\log 2\] \[\implies 5\log\left(4y^2\right)=-\log\left(16z^2\right)\implies 5\log\left(4y^2\right)+\log\left(16z^2\right)=0\] \[\implies \left(4y^2\right)^5\cdot 16z^2=1\implies 16384y^{10}z^2=1\implies y^{10}z^2=\frac{1}{16384}\implies y^5z=\frac{1}{128}\]
Thus $xy^5z=2^{-\frac{1}{6}-7}=2^{-\frac{43}{6}}$ and $43+6=\boxed{049}$

Answer:  049
Problem: Let $x,$ $y,$ and $z$ be positive real numbers that satisfy \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.\] The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution: Since we are given that $xy^5z = 2^{-p/q}$ , we may assume that $x, y$ , and $z$ are all powers of two.  We shall thus let $x = 2^X$ $y = 2^Y$ , and $z = 2^Z$ .  Let $a = \log_{2^X}(2^{Y+1})$ .  From this we get the system of equations: \[\] $(1)$ \[a = \log_{2^X}(2^{Y+1}) \Rightarrow aX = Y + 1\] $(2)$ \[a = \log_{2^{X + 1}}(2^{Z + 2}) \Rightarrow aX + a = Z + 2\] $(3)$ \[2a = \log_{2^{4X + 1}}(2^{Y + Z + 3}) \Rightarrow 8aX + 2a = Y + Z + 3\]
Plugging equation $(1)$ into equation $(2)$ yields $Y + a = Z + 1$ .  Plugging equation $(1)$ into equation $(3)$ and simplifying yields $7Y + 2a + 6 = Z + 1$ , and substituting $Y + a$ for $Z + 1$ and simplifying yields $Y + 1 = \frac{-a}{6}$ .  But $Y + 1 = aX$ , so $aX =  \frac{-a}{6}$ , so $X = \frac{-1}{6}$
Knowing this, we may substitute $\frac{-1}{6}$ for $X$ in equations $(1)$ and $(2)$ , yielding $\frac{-a}{6} = Y + 1$ and $\frac{5a}{6} = Z + 2$ .  Thus, we have that $-5(Y + 1) = Z + 2 \rightarrow 5Y + Z = -7$ .  We are looking for $xy^5z = 2^{X+ 5Y + Z}$ $X = \frac{-1}{6}$ and $5Y + Z = -7$ , so $xy^5z = 2^{-43/6} = \frac{1}{2^{43/6}}$ .  The answer is $43+6=\boxed{049}$

Answer:  049
Problem: Let $x,$ $y,$ and $z$ be positive real numbers that satisfy \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.\] The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution: We know that \[\frac{\log (4y^2)}{\log (x)} = \frac{\log (16z^2)}{\log (2x)} = \frac{2\log (8yz)}{\log (2x^2)}\] By the Mediant theorem
Substituting into the original equation yields us $\frac{2\log (8yz)}{\log (2x^2)} = \frac{\log (8yz)}{\log (2x^4)} \Rightarrow 2\log (2x^4) = \log (2x^2) \Rightarrow x=2^{-1/6}.$ For some constant $C\not= 0,$ Let $2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) = C.$ Then, we obtain the system of equations \[y=2^{-13C/12}\] \[z=2^{-19C/12}\] \[8yz=2^{C/3}.\]
Adding the first two equations and subtracting the third, we find $C=1.$ Thus, \[xy^5z=2^{-1/6} \cdot 2^{-65/12} \cdot 2^{-19/12}=2^{-43/6} \Rightarrow p+q=\boxed{049}.\]

Answer:  049
Problem: Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$ . Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$ , where $x$ is in $\mathcal{S}$ . In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$

Solution: It is apparent that for a perfect square $s^2$ to satisfy the constraints, we must have $s^2 - 256 = 1000n$ or $(s+16)(s-16) = 1000n.$ Now in order for $(s+16)(s-16)$ to be a multiple of $1000,$ at least one of $s+16$ and $s-16$ must be a multiple of $5,$ and since $s+16 \not\equiv s-16 \pmod{5},$ one term must have all the factors of $5$ and thus must be a multiple of $125.$ Furthermore, each of $s+16$ and $s-16$ must have at least two factors of $2,$ since otherwise $(s+16)(s-16)$ could not possibly be divisible by $8.$ So therefore the conditions are satisfied if either $s+16$ or $s-16$ is divisible by $500,$ or equivalently if $s = 500n \pm 16.$ Counting up from $n=0$ to $n=5,$ we see that the tenth value of $s$ is $500 \cdot 5 - 16 = 2484$ and thus the corresponding element in $\mathcal{T}$ is $\frac{2484^2 - 256}{1000} = 6170 \rightarrow \boxed{170.}$

Answer:  170.
Problem: Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$ . Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$ , where $x$ is in $\mathcal{S}$ . In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$

Solution: Notice that is $16^2=256$ $1016^2$ ends in $256$ . In general, if $x^2$ ends in $256$ $(x+1000)^2=x^2+2000x+1000000$ ends in 256 because $1000|2000x$ and $1000|2000000$ . It is clear that we want all numbers whose squares end in $256$ that are less than $1000$
Firstly, we know the number has to end in a $4$ or a $6$ . Let's look at the ones ending in $6$
Assume that the second digit of the three digit number is $0$ . We find that the last $3$ digits of $\overline{a06}^2$ is in the form $12a \cdot 100 + 3 \cdot 10 + 6$ . However, the last two digits need to be a $56$ . Thus, similarly trying all numbers from $0$ to $10$ , we see that only 1 for the middle digit works. Carrying out the multiplication, we see that the last $3$ digits of $\overline{a06}^2$ is in the form $(12a + 2) \cdot 100 + 5 \cdot 10 + 6$ . We want $(12a + 2)\pmod{10}$ to be equal to $2$ . Thus, we see that a is $0$ or $5$ . Thus, the numbers that work in this case are $016$ and $516$
Next, let's look at the ones ending in $4$ . Carrying out a similar technique as above, we see that the last $3$ digits of $\overline{a84}^2$ is in the form $((8a+10) \cdot 100+ 5 \cdot 10 + 6$ . We want $(8a + 10)\pmod{10}$ to be equal to $2$ . We see that only $4$ and $9$ work. Thus, we see that only $484$ and $984$ work.
We order these numbers to get $16$ $484$ $516$ $984$ ... We want the $10th$ number in order which is $2484^2 = 6\boxed{170}256$

Answer:  170
Problem: Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$ . Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$ , where $x$ is in $\mathcal{S}$ . In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$

Solution: The condition implies $x^2\equiv 256 \pmod{1000}$ . Rearranging and factoring, \[(x-16)(x+16)\equiv 0\pmod {1000}.\] This can be expressed with the system of congruences \[\begin{cases}  (x-16)(x+16)\equiv 0\pmod{125} \\ (x-16)(x+16)\equiv 0\pmod{8} \end{cases}\] Observe that $x\equiv {109} \pmod {125}$ or $x\equiv{16}\pmod {125}$ . Similarly, it can be seen that $x\equiv{0}\pmod{8}$ or $x\equiv{4}\pmod{8}$ . By CRT, there are four solutions to this modulo $1000$ (one for each case e.g. $x\equiv{109}$ and $x\equiv{0}$ or $x\equiv{125}$ and $x\equiv{4}$ . These solutions are (working modulo $1000$ \[\begin{cases}  x=16 \\ x=484 \\ x= 516 \\ x=984 \end{cases}\] The tenth solution is $x=2484,$ which gives an answer of $\boxed{170}$

Answer:  170
Problem: Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$ . Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$ , where $x$ is in $\mathcal{S}$ . In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$

Solution: An element in S can be represented by $y^2 = 1000a + 256$ , where $y^2$ is the element in S. Since the right hand side must be even, we let $y = 2y_1$ and substitute to get $y_1^2 = 250a + 64$ . However, the right hand side is still even, so we make the substitution $y_1 = 2y_2$ to get $y_2^2 = 125a/2 + 16$ . Because both sides must be an integer, we know that $a = 2a_1$ for some integer $a_1$ . Our equation then becomes $y_2^2 = 125a_1 + 16$ , and we can simplify no further.
Rearranging terms, we get $y_2^2 - 16 = 125a_1$ , whence difference of squares gives $(y_2 + 4)(y_2 - 4) = 125a_1$ . Note that this equation tells us that one of $y_2 + 4$ and $y_2 - 4$ contains a nonnegative multiple of $125$ . Hence, listing out the smallest possible values of $y_2$ , we have $y_2 = 4, 121, 129, \cdots, 621$ . The tenth term is $621$ , whence $y = 4y_2 = 2484$ . The desired result can then be calculated to be $\boxed{170}$ . - Spacesam

Answer:  170
Problem: Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$ . Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$ , where $x$ is in $\mathcal{S}$ . In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$

Solution: From the conditions, we can let every element in $\mathcal{S}$ be written as $y^2=1000x+256$ , where $x$ and $y$ are integers. Since there are no restrictions on $y$ , we let $y_1$ be equal to $y+16$ $y-16$ works as well). Then the $256$ cancels out and we're left with \[y_1^2+32y_1=1000x\] which can be factored as \[y_1(y_1+32)=1000x\] Since the RHS is even, $y_1$ must be even, so we let $y_1=2y_2$ , to get \[y_2(y_2+16)=250x\] Again, because the RHS is even, the LHS must be even too, so substituting $y_3=\frac{1}{2}y_2$ we have \[y_3(y_3+8)=125\cdot\frac{1}{2}x\] Since the LHS is an integer, the RHS must thus be an integer, so substituting $x=2x_1$ we get \[y_3(y_3+8)=125x_1\] Then we can do casework on the values of $y_3$ , as only one of $y_3$ and $y_3+8$ can be multiples of $125$
Case 1 $125|y_3$
Since we're trying to find the values of $x_1$ , we can let $y_4=\frac{1}{125}y_3$ , to get \[x_1=y_4(125y_4+8)\] or \[x=2y_4(125y_4+8)\]
Case 2 $125|y_3+8$
Similar to Case 1, only the equation is \[x=2y_4(125y_4-8)\]
In whole, the values of $x$ (i.e. the elements in $\mathcal{T}$ ) are of the form \[x=2k(125k\pm8)\] where $k$ is any integer. It can easily be seen that if $k<0$ , then $x$ is negative, thus $k\geq0$ . Also, note that when $k=0$ , there is only one value, because one of the factors is $0$ $k$ ). Thus the $10^{th}$ smallest number in the set $\mathcal{T}$ is when the $\pm$ sign takes the minus side and $k=5$ , giving $6170$ , so the answer is $\boxed{170}$

Answer:  170
Problem: Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$

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

Answer:  018
Problem: Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$

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

Answer:  018
Problem: Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$

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

Answer:  018