Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$

Solution: First scale down the whole cube by $12$ . Let point $P$ have coordinates $(x, y, z)$ , point $A$ have coordinates $(0, 0, 0)$ , and $s$ be the side length. Then we have the equations \begin{align*} (s-x)^2+y^2+z^2&=\left(5\sqrt{10}\right)^2, \\ x^2+(s-y)^2+z^2&=\left(5\sqrt{5}\right)^2, \\ x^2+y^2+(s-z)^2&=\left(10\sqrt{2}\right)^2, \\ (s-x)^2+(s-y)^2+(s-z)^2&=\left(3\sqrt{7}\right)^2. \end{align*} These simplify into \begin{align*} s^2+x^2+y^2+z^2-2sx&=250, \\ s^2+x^2+y^2+z^2-2sy&=125, \\ s^2+x^2+y^2+z^2-2sz&=200, \\ 3s^2-2s(x+y+z)+x^2+y^2+z^2&=63. \end{align*} Adding the first three equations together, we get $3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575$ .
Subtracting this from the fourth equation, we get $2(x^2+y^2+z^2)=512$ , so $x^2+y^2+z^2=256$ . This means $PA=16$ . However, we scaled down everything by $12$ so our answer is $16*12=\boxed{192}$

Answer:  192
Problem: Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$

Solution: Once the equations for the distance between point P and the vertices of the cube have been written, we can add the first, second, and third to receive, \[2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.\] Subtracting the fourth equation gives \begin{align*} 2(x^2 + y^2 + z^2) &= 575 - 63 \\ x^2 + y^2 + z^2 &= 256 \\ \sqrt{x^2 + y^2 + z^2} &= 16. \end{align*} Since point $A = (0,0,0), PA = 16$ , and since we scaled the answer is $16 \cdot 12 = \boxed{192}$

Answer:  192
Problem: Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$

Solution: Let $E$ be the vertex of the cube such that $ABED$ is a square.
Using the British Flag Theorem , we can easily show that \[PA^2 + PE^2 = PB^2 + PD^2\] and \[PA^2 + PG^2 = PC^2 + PE^2\] Hence, by adding the two equations together, we get $2PA^2 + PG^2 = PB^2 + PC^2 + PD^2$ . Substituting in the values we know, we get $2PA^2 + 7\cdot 36^2 =10\cdot60^2 + 5\cdot 60^2 + 2\cdot 120^2$
Thus, we can solve for $PA$ , which ends up being $\boxed{192}$

Answer:  192
Problem: Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$

Solution: For all points $X$ in space, define the function $f:\mathbb{R}^{3}\rightarrow\mathbb{R}$ by $f(X)=PX^{2}-GX^{2}$ . Then $f$ is linear; let $O=\frac{2A+G}{3}$ be the center of $\triangle BCD$ . Then since $f$ is linear, \begin{align*} 3f(O)=f(B)+f(C)+f(D)&=2f(A)+f(G) \\  \left(PB^{2}-GB^{2}\right)+\left(PC^{2}-GC^{2}\right)+\left(PD^{2}-GD^{2}\right)&=2\left(PA^{2}-GA^{2}\right)+PG^{2} \\ \left(60\sqrt{10}\right)^{2}-2x^{2}+\left(60\sqrt{5}\right)^{2}-2x^{2}+\left(120\sqrt{2}\right)^{2}-2x^{2}&=2PA^{2}-2\cdot 3x^{2}+\left(36\sqrt{7}\right)^{2}, \end{align*} where $x$ denotes the side length of the cube. Thus \begin{align*}  36\text000+18\text000+28\text800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\  82\text800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\  73\text728&=2PA^{2} \\  36\text864&=PA^{2} \\  PA&=\boxed{192}

Answer:  192
Problem: Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\]

Solution: It is trivial that the maximum value of $\sin \theta$ is $1$ , is achieved at $\theta = \frac{\pi}{2}+2k\pi$ for some integer $k$
This implies that $\sin(mx) = \sin(nx) = 1$ , and that $mx = \frac{\pi}{2}+2a\pi$ and $nx = \frac{\pi}{2}+2b\pi$ , for integers $a, b$
Taking their ratio, we have \[\frac{mx}{nx} = \frac{\frac{\pi}{2}+2a\pi}{\frac{\pi}{2}+2b\pi} \implies \frac{m}{n} = \frac{4a + 1}{4b + 1} \implies \frac{m}{4a + 1} = \frac{n}{4b + 1} = k.\] It remains to find all $m, n$ that satisfy this equation.
If $k = 1$ , then $m \equiv n \equiv 1 \pmod 4$ . This corresponds to choosing two elements from the set $\{1, 5, 9, 13, 17, 21, 25, 29\}$ . There are $\binom 82$ ways to do so.
If $k < 1$ , by multiplying $m$ and $n$ by the same constant $c = \frac{1}{k}$ , we have that $mc \equiv nc \equiv 1 \pmod 4$ . Then either $m \equiv n \equiv 1 \pmod 4$ , or $m \equiv n \equiv 3 \pmod 4$ . But the first case was already counted, so we don't need to consider that case. The other case corresponds to choosing two numbers from the set $\{3, 7, 11, 15, 19, 23, 27\}$ . There are $\binom 72$ ways here. (This argument seems to have a logical flaw)
Finally, if $k > 1$ , note that $k$ must be an integer. This means that $m, n$ belong to the set $\{k, 5k, 9k, \dots\}$ , or $\{3k, 7k, 11k, \dots\}$ . Taking casework on $k$ , we get the sets $\{2, 10, 18, 26\}, \{6, 14, 22, 30\}, \{4, 20\}, \{12, 28\}$ . Some sets have been omitted; this is because they were counted in the other cases already. This sums to $\binom 42 + \binom 42 + \binom 22 + \binom 22$
In total, there are $\binom 82 + \binom 72 + \binom 42 + \binom 42 + \binom 22 + \binom 22 = \boxed{063}$ pairs of $(m, n)$

Answer:  063
Problem: Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\]

Solution: In order for $\sin(mx) + \sin(nx) = 2$ $\sin(mx) = \sin(nx) = 1$
This happens when $mx \equiv nx \equiv \frac{\pi}{2} ($ mod $2\pi).$
This means that $mx = \frac{\pi}{2} + 2\pi\alpha$ and $nx = \frac{\pi}{2} + 2\pi\beta$ for any integers $\alpha$ and $\beta$
As in Solution 1, take the ratio of the two equations: \[\frac{mx}{nx} = \frac{\frac{\pi}{2}+2\pi\alpha}{\frac{\pi}{2}+2\pi\beta} \implies \frac{m}{n} = \frac{\frac{1}{2}+2\alpha}{\frac{1}{2}+2\beta} \implies \frac{m}{n} = \frac{4\alpha+1}{4\beta+1}\]
Now notice that the numerator and denominator of $\frac{4\alpha+1}{4\beta+1}$ are both odd, which means that $m$ and $n$ have the same power of two (the powers of 2 cancel out).
Let the common power be $p$ : then $m = 2^p\cdot a$ , and $n = 2^p\cdot b$ where $a$ and $b$ are integers between 1 and 30.
We can now rewrite the equation: \[\frac{2^p\cdot a}{2^p\cdot b} = \frac{4\alpha+1}{4\beta+1} \implies \frac{a}{b} = \frac{4\alpha+1}{4\beta+1}\]
Now it is easy to tell that $a \equiv 1 ($ mod $4)$ and $b \equiv 1 ($ mod $4)$ . However, there is another case: that
$a \equiv 3 ($ mod $4)$ and $b \equiv 3 ($ mod $4)$ . This is because multiplying both $4\alpha+1$ and $4\beta+1$ by $-1$ will not change the fraction, but each congruence will be changed to $-1 ($ mod $4) \equiv 3 ($ mod $4)$
From the first set of congruences, we find that $a$ and $b$ can be two of $\{1, 5, 9, \ldots, 29\}$
From the second set of congruences, we find that $a$ and $b$ can be two of $\{3, 7, 11, \ldots, 27\}$
Now all we have to do is multiply by $2^p$ to get back to $m$ and $n$ . 
Let’s organize the solutions in order of increasing values of $p$ , keeping in mind that $m$ and $n$ are bounded between 1 and 30.
For $p = 0$ we get $\{1, 5, 9, \ldots, 29\}, \{3, 7, 11, \ldots, 27\}$
For $p = 1$ we get $\{2, 10, 18, 26\}, \{6, 14, 22, 30\}$
For $p = 2$ we get $\{4, 20\}, \{12, 28\}$
Note that $16\mid{a}$ since $m$ will cancel out a factor of 4 from $a$ , and $\frac{a}{m}$ must contain a factor of 4. Again, $1-4X$ will never contribute a factor of 2. Simply inspecting, we see two feasible values for $a$ and $m$ such that $a+m\leq30$
If we increase the value of $p$ more, there will be less than two integers in our sets, so we are done there.
There are 8 numbers in the first set, 7 in the second, 4 in the third, 4 in the fourth, 2 in the fifth, and 2 in the sixth.
In each of these sets we can choose 2 numbers to be $m$ and $n$ and then assign them in increasing order. Thus there are:
\[\dbinom{8}{2}+\dbinom{7}{2}+\dbinom{4}{2}+\dbinom{4}{2}+\dbinom{2}{2}+\dbinom{2}{2} = 28+21+6+6+1+1 = \boxed{063}\] possible pairs $(m,n)$ that satisfy the conditions.

Answer:  063
Problem: Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\]

Solution: We know that the range of sine is between $-1$ and $1$ , inclusive.
Thus, the only way for the sum to be $2$ is for $\sin(mx)=\sin(nx)=1$
Note that $\sin(90+360k)=1$
Assuming $mx$ and $nx$ are both positive, $m$ and $n$ could be $1,5,9,13,17,21,25,29$ . There are $8$ ways, so $\dbinom{8}{2}$
If both are negative, $m$ and $n$ could be $3,7,11,15,19,23,27$ . There are $7$ ways, so $\dbinom{7}{2}$
However, the pair $(1,5)$ could also be $(2, 10)$ and so on. The same goes for some other pairs.
In total there are $14$ of these extra pairs.
The answer is $28+21+14 = \boxed{063}$

Answer:  063
Problem: Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\]

Solution: The equation implies that $\sin(mx)=\sin(nx)=1$ . Therefore, we can write $mx$ as $2{\pi}k_1+\frac{\pi}{2}$ and $nx$ as $2{\pi}k_2+\frac{\pi}{2}$ for integers $k_1$ and $k_2$ . Then, $\frac{mx}{nx}=\frac{m}{n}=\frac{2k_1+\frac{1}{2}}{2k_2+\frac{1}{2}}$ . Cross multiplying, we get $m\cdot{(2k_2+\frac{1}{2})}=n\cdot{(2k_1+\frac{1}{2})} \Longrightarrow 4k_2m-4k_1n=n-m$ . Let $n-m=a$ so the equation becomes $4(m(k_2-k_1)+k_1a)=a$ . Let $k_2-k_1=X$ and $k_1=Y$ , then the equation becomes $a=4Ym+4Xa \Longrightarrow \frac{a(1-4X)}{m}=4Y$ . Note that $X$ and $Y$ can vary accordingly, and $4\mid{a}$ . Next, we do casework on $m\pmod{4}$
If $m\equiv 1\pmod{4}$
Once $a$ and $m$ are determined, $n$ is determined, so $a+m\leq30$ $a\in \{4,8,12,\dots,28\}$ and $m\in \{1,5,9,\dots,29\}$ . Therefore, there are $\sum_{i=1}^{7}{i}=28$ ways for this case such that $a+m\leq30$
If $m\equiv 3\pmod{4}$
$a\in \{4,8,12,\dots,28\}$ and $m\in \{3,7,11,\dots,27\}$ . Therefore, there are $\sum_{i=1}^{6}{i}=21$ ways such that $a+m\leq30$
If $m\equiv 2\pmod{4}$
Note that $8\mid{a}$ since $m$ in this case will have a factor of 2, which will cancel out a factor of 2 in $a$ , and we need the left hand side to divide 4. Also, $1-4X\equiv 1\pmod{4}$ so it is odd and will therefore never contribute a factor of 2. $a\in \{8,16,24\}$ and $m\in \{2,6,10,\dots,30\}$ . Following the condition $a+m\leq30$ , we conclude that there are $6+4+2=12$ ways for this case.
If $m\equiv 0\pmod{4}$
Adding all the cases up, we obtain $28+21+12+2=\boxed{063}$

Answer:  063
Problem: Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\] has $12$ distinct real solutions.

Solution: We take cases for the outermost absolute value, then rearrange: \[\left|20|x|-x^2\right|=c\pm21.\] Let $f(x)=\left|20|x|-x^2\right|.$ We rewrite $f(x)$ as a piecewise function without using absolute values: \[f(x) = \begin{cases} \left|-20x-x^2\right| & \mathrm{if} \ x \le 0  \begin{cases} 20x+x^2 & \mathrm{if} \ x\le-20 \\ -20x-x^2 & \mathrm{if} \ -20<x\leq0 \end{cases} \\ \left|20x-x^2\right| & \mathrm{if} \ x > 0 \begin{cases} 20x-x^2 & \mathrm{if} \ 0<x\leq20 \\ -20x+x^2 & \mathrm{if} \ x>20 \end{cases} \end{cases}.\] We graph $y=f(x)$ with all extremum points labeled, as shown below. The fact that $f(x)$ is an even function ( $f(x)=f(-x)$ holds for all real numbers $x,$ so the graph of $y=f(x)$ is symmetric about the $y$ -axis) should facilitate the process of graphing. [asy] /* Made by MRENTHUSIASM */ size(1200,300);   real xMin = -65; real xMax = 65; real yMin = -50; real yMax = 125;  draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2));  real f(real x) { return abs(20*abs(x)-x^2); } real g(real x) { return 21; } real h(real x) { return -21; } draw(graph(f,-25,25),red,"$y=\left|20|x|-x^2\right|$"); draw(graph(g,-65,65),blue,"$y=\pm21$"); draw(graph(h,-65,65),blue);  pair A[]; A[0] = (-20,0); A[1] = (-10,100); A[2] = (0,0); A[3] = (10,100); A[4] = (20,0);  for(int i = 0; i <= 4; ++i) { dot(A[i],red+linewidth(4.5));  }  label("$(-20,0)$",A[0],(-1.5,-1.5),red,UnFill); label("$(-10,100)$",A[1],(-1.5,1.5),red); label("$(0,0)$",A[2],(0,-1.5),red,UnFill); label("$(10,100)$",A[3],(1.5,1.5),red); label("$(20,0)$",A[4],(1.5,-1.5),red,UnFill);  add(legend(),point(E),40E,UnFill); [/asy] Since $f(x)=c\pm21$ has $12$ distinct real solutions, it is clear that each case has $6$ distinct real solutions geometrically. We shift the graphs of $y=\pm21$ up $c$ units, where $c\geq0:$
Taking the intersection of these two cases gives $21<c<79,$ from which there are $79-21-1=\boxed{057}$ such integers $c.$

Answer:  057
Problem: Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\] has $12$ distinct real solutions.

Solution: Graph $y=|20|x|-x^2|$ (If you are having trouble, look at the description in the next two lines and/or the diagram in Solution 1). Notice that we want this to be equal to $c-21$ and $c+21$
We see that from left to right, the graph first dips from very positive to $0$ at $x=-20$ , then rebounds up to $100$ at $x=-10$ , then falls back down to $0$ at $x=0$
The positive $x$ are symmetric, so the graph re-ascends to $100$ at $x=10$ , falls back to $0$ at $x=10$ , and rises to arbitrarily large values afterwards.
Now we analyze the $y$ (varied by $c$ ) values. At $y=k<0$ , we will have no solutions, as the line $y=k$ will have no intersections with our graph.
At $y=0$ , we will have exactly $3$ solutions for the three zeroes.
At $y=n$ for any $n$ strictly between $0$ and $100$ , we will have exactly $6$ solutions.
At $y=100$ , we will have $4$ solutions, because local maxima are reached at $x= \pm 10$
At $y=m>100$ , we will have exactly $2$ solutions.
To get $12$ distinct solutions for $y=|20|x|-x^2|=c \pm 21$ , both $c +21$ and $c-21$ must produce $6$ solutions.
Thus $0<c-21$ and $c+21<100$ , so $c \in \{ 22, 23, \dots , 77, 78 \}$ is required.
It is easy to verify that all of these choices of $c$ produce $12$ distinct solutions (none overlap), so our answer is $\boxed{057}$

Answer:  057
Problem: Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\] has $12$ distinct real solutions.

Solution: Let $y = |x|.$ Then the equation becomes $\left|\left|20y-y^2\right|-c\right| = 21$ , or $\left|y^2-20y\right| = c \pm 21$ . Note that since $y = |x|$ $y$ is nonnegative, so we only care about nonnegative solutions in $y$ . Notice that each positive solution in $y$ gives two solutions in $x$ $x = \pm y$ ), whereas if $y = 0$ is a solution, this only gives one solution in $x$ $x = 0$ . Since the total number of solutions in $x$ is even, $y = 0$ must not be a solution. Hence, we require that $\left|y^2-20y\right| = c \pm 21$ has exactly $6$ positive solutions and is not solved by $y = 0.$
If $c < 21$ , then $c - 21$ is negative, and therefore cannot be the absolute value of $y^2 - 20y$ . This means the equation's only solutions are in $\left|y^2-20y\right| = c + 21$ . There is no way for this equation to have $6$ solutions, since the quadratic $y^2-20y$ can only take on each of the two values $\pm(c + 21)$ at most twice, yielding at most $4$ solutions. Hence, $c \ge 21$ $c$ also can't equal $21$ , since this would mean $y = 0$ would solve the equation. Hence, $c > 21.$
At this point, the equation $y^2-20y = c \pm 21$ will always have exactly $2$ positive solutions, since $y^2-20y$ takes on each positive value exactly once when $y$ is restricted to positive values (graph it to see this), and $c \pm 21$ are both positive. Therefore, we just need $y^2-20y = -(c \pm 21)$ to have the remaining $4$ solutions exactly. This means the horizontal lines at $-(c \pm 21)$ each intersect the parabola $y^2 - 20y$ in two places. This occurs when the two lines are above the parabola's vertex $(10,-100)$ . Hence we have \begin{align*} -(c + 21) &> -100 \\ c + 21 &< 100 \\ c &< 79. \end{align*} Hence, the integers $c$ satisfying the conditions are those satisfying $21 < c < 79.$ There are $\boxed{057}$ such integers.

Answer:  057
Problem: Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\] has $12$ distinct real solutions.

Solution: Removing the absolute value bars from the equation successively, we get \begin{align*} \left|\left|20|x|-x^2\right|-c\right|&=21 \\ \left|20|x|-x^2\right|&= c \pm21 \\ 20|x|-x^2 &= \pm c \pm 21 \\ x^2 \pm 20x \pm c \pm21 &= 0. \end{align*} The discriminant of this equation is \[\sqrt{400-4(\pm c \pm 21)}.\] Equating the discriminant to $0$ , we see that there will be two distinct solutions to each of the possible quadratics above only in the interval $-79 < c < 79$ . However, the number of zeros the equation $ax^2+b|x|+k$ has is determined by where $ax^2+bx+k$ and $ax^2-bx+k$ intersect, namely at $(0,k)$ . When $k<0$ $a>0$ $ax^2+b|x|+k$ will have only $2$ solutions, and when $k>0$ $a>0$ , then there will be $4$ real solutions, if they exist at all.
In order to have $12$ solutions here, we thus need to ensure $-c+21<0$ , so that exactly $2$ out of the $4$ possible equations of the form $ax^2+b|x|+k$ given above have y-intercepts below $0$ and only $2$ real solutions, while the remaining $2$ equations have $4$ solutions. This occurs when $c>21$ , so our final bounds are $21<c<79$ , giving us $\boxed{057}$ valid values of $c$

Answer:  057
Problem: Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$

Solution: Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to $\overleftrightarrow{BC}, \overleftrightarrow{CD},$ and $\overleftrightarrow{BD},$ respectively. Next, let $H$ be the intersection of $\overline{AF}$ and $\overline{BD}.$
We set $AB=x$ and $AH=y,$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); dot("$A$",A,1.5*NW,linewidth(4)); dot("$B$",B,1.5*NE,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*S,linewidth(4)); dot("$G$",G,SE,linewidth(4)); dot("$H$",H,SE,linewidth(4)); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label("$10$",midpoint(A--G),1.5*(1,0)); label("$15$",midpoint(A--E),1.5*N); Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); label("$x$",midpoint(A--B),N); label("$y$",midpoint(A--H),W); [/asy] From here, we obtain $HF=18-y$ by segment subtraction, and $BG=\sqrt{x^2-10^2}$ and $HG=\sqrt{y^2-10^2}$ by the Pythagorean Theorem.
Since $\angle ABG$ and $\angle HAG$ are both complementary to $\angle AHB,$ we have $\angle ABG = \angle HAG,$ from which $\triangle ABG \sim \triangle HAG$ by AA. It follows that $\frac{BG}{AG}=\frac{AG}{HG},$ so $BG\cdot HG=AG^2,$ or \[\sqrt{x^2-10^2}\cdot\sqrt{y^2-10^2}=10^2. \hspace{10mm} (1)\] Since $\angle AHB = \angle FHD$ by vertical angles, we have $\triangle AHB \sim \triangle FHD$ by AA, with the ratio of similitude $\frac{AH}{FH}=\frac{BA}{DF}.$ It follows that $DF=BA\cdot\frac{FH}{AH}=x\cdot\frac{18-y}{y}.$
Since $\angle EBA = \angle ECD = \angle FDA$ by angle chasing, we have $\triangle EBA \sim \triangle FDA$ by AA, with the ratio of similitude $\frac{EA}{FA}=\frac{BA}{DA}.$ It follows that $DA=BA\cdot\frac{FA}{EA}=x\cdot\frac{18}{15}=\frac{6}{5}x.$
By the Pythagorean Theorem on right $\triangle ADF,$ we have $DF^2+AF^2=AD^2,$ or \[\left(x\cdot\frac{18-y}{y}\right)^2+18^2=\left(\frac{6}{5}x\right)^2. \hspace{7mm} (2)\] Solving this system of equations ( $(1)$ and $(2)$ ), we get $x=\frac{45\sqrt2}{4}$ and $y=\frac{90}{7},$ so $AB=x=\frac{45\sqrt2}{4}$ and $CD=AB+2DF=x+2\left(x\cdot\frac{18-y}{y}\right)=\frac{81\sqrt2}{4}.$ Finally, the area of $ABCD$ is \[K=\frac{AB+CD}{2}\cdot AF=\frac{567\sqrt2}{2},\] from which $\sqrt2 \cdot K=\boxed{567}.$

Answer:  567
Problem: Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$

Solution: First, draw the diagram. Then, notice that since $ABCD$ is isosceles, $\Delta ABD \cong \Delta BAC$ , and the length of the altitude from $B$ to $AC$ is also $10$ . Let the foot of this altitude be $F$ , and let the foot of the altitude from $A$ to $BC$ be denoted as $E$ . Then, $\Delta BCF \sim \Delta ACE$ . So, $\frac{BC}{AC} = \frac{BF}{AE} = \frac{2}{3}$ . Now, notice that $[ABC] = \frac{10 \cdot AC} {2} = \frac{AB \cdot 18}{2} \implies AC = \frac{9 \cdot AB}{5}$ , where $[ABC]$ denotes the area of triangle $ABC$ . Letting $AB = x$ , this equality becomes $AC = \frac{9x}{5}$ . Also, from $\frac{BC}{AC} = \frac{2}{3}$ , we have $BC = \frac{6x}{5}$ . Now, by the Pythagorean theorem on triangles $ABF$ and $CBF$ , we have $AF = \sqrt{x^{2}-100}$ and $CF = \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}$ . Notice that $AC = AF + CF$ , so $\frac{9x}{5} = \sqrt{x^{2}-100} + \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}$ . Squaring both sides of the equation once, moving $x^{2}-100$ and $\left( \frac{6x}{5} \right) ^{2}-100$ to the right, dividing both sides by $2$ , and squaring the equation once more, we are left with $\frac{32x^{4}}{25} = 324x^{2}$ . Dividing both sides by $x^{2}$ (since we know $x$ is positive), we are left with $\frac{32x^{2}}{25} = 324$ . Solving for $x$ gives us $x = \frac{45}{2\sqrt{2}}$
Now, let the foot of the perpendicular from $A$ to $CD$ be $G$ . Then let $DG = y$ . Let the foot of the perpendicular from $B$ to $CD$ be $H$ . Then, $CH$ is also equal to $y$ . Notice that $ABHG$ is a rectangle, so $GH = x$ . Now, we have $CG = GH + CH = x + y$ . By the Pythagorean theorem applied to $\Delta AGC$ , we have $(x+y)^{2}+18^{2}= \left( \frac{9x}{5} \right) ^{2}$ . We know that $\frac{9x}{5} = \frac{9}{5} \cdot \frac{45}{2\sqrt{2}} = \frac{81}{2\sqrt{2}}$ , so we can plug this into this equation. Solving for $x+y$ , we get $x+y=\frac{63}{2\sqrt{2}}$
Finally, to find $[ABCD]$ , we use the formula for the area of a trapezoid: $K = [ABCD] = \frac{b_{1}+b_{2}}{2} \cdot h = \frac{AB+CD}{2} \cdot 18 = \frac{x+(CG+DG)}{2} \cdot 18 = \frac{2x+2y}{2} \cdot 18 = (x+y) \cdot 18 = \frac{63}{2\sqrt{2}} \cdot 18 = \frac{567}{\sqrt{2}}$ . The problem asks us for $K \cdot \sqrt{2}$ , which comes out to be $\boxed{567}$

Answer:  567