Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of triangle $CDM$ may be expressed as $\frac {m\sqrt {n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m + n + p.$
Solution 1
We use the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) on $ABC$ to determine that $AB=25.$
Let $N$ be the orthogonal projection from $C$ to $AB.$ Thus, $[CDM]=\frac{(DM)(MN)} {2}$, $MN=BN-BM$, and $[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.$
From the third equation, we get 
$CN=\frac{168} {25}.$
By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) in $\Delta BCN,$ we have 
$BN=\sqrt{\left(\frac{24 \cdot 25} {25}\right)^2-\left(\frac{24 \cdot 7} {25}\right)^2}=\frac{24} {25}\sqrt{25^2-7^2}=\frac{576} {25}.$
Thus, 
$MN=\frac{576} {25}-\frac{25} {2}=\frac{527} {50}.$
In $\Delta ADM$, we use the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) to get 
$DM=\sqrt{15^2-\left(\frac{25} {2}\right)^2}=\frac{5} {2} \sqrt{11}.$
Thus, 
$[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}= \frac{527\sqrt{11}} {40}.$
Hence, the answer is $527+11+40=\boxed{578}.$
~ minor edits by kundusne000

Solution 2
By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) in $\Delta AMD$, we get $DM=\frac{5\sqrt{11}} {2}$. Since $ABC$ is a right triangle, $M$ is the circumcenter and thus, $CM=\frac{25} {2}$. We let $\angle CMD=\theta$. By the [Law of Cosines](https://artofproblemsolving.com/wiki/index.php/Law_of_Cosines),
$24^2 = 2 \cdot (12.5)^2-2 \cdot (12.5)^2 * \cos (90+\theta).$
It follows that $\sin \theta = \frac{527} {625}$. Thus,
$[CMD]=\frac{1} {2} (12.5) \left(\frac{5\sqrt{11}} {2}\right)\left(\frac{527} {625}\right)=\frac{527\sqrt{11}} {40}$.

Solution 3
Suppose $ABC$ is plotted on the [cartesian plane](https://artofproblemsolving.com/wiki/index.php/Cartesian_plane) with $C$ at $(0,0)$, $A$ at $(0,7)$, and $B$ at $(24,0)$.
Then $M$ is at $(12,3.5)$. Since $\Delta ABD$ is isosceles, $MD$ is perpendicular to $AM$, and since $AM=12.5$ and $AD=15, MD=2.5\sqrt{11}$.
The slope of $AM$ is $-\frac{7}{24}$ so the slope of $MD$ is $\frac{24}{7}$.
Draw a vertical line through $M$ and a horizontal line through $D$. Suppose these two lines meet at $X$. then $MX=\frac{24}{7}DX$ so $MD=\frac{25}{7}DX=\frac{25}{24}MD$ by the pythagorean theorem.
So $MX=2.4\sqrt{11}$ and $DX=.7\sqrt{11}$ so the coordinates of D are $(12-.7\sqrt{11},\, 3.5-2.4\sqrt{11})$.
Since we know the coordinates of each of the vertices of $\Delta CMD$, we can apply the [Shoelace Theorem](https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem) to find the area of $\Delta CMD, \frac{527 \sqrt{11}}{40}$.
~ minor edit by kundusne000

Solution 4
Let $E$ be the intersection of lines $BC$ and $DM$.  Since triangles $\Delta CME$ and $\Delta CMD$ share a side and height, the area of $\Delta CDM$ is equal to $\frac{DM}{EM}$ times the area of $\Delta CME$.
By AA similarity, $\Delta EMB$ is similar to $\Delta ACB$, $\frac{EM}{AC}=\frac{MB}{CB}$. Solving yields $EM=\frac{175}{48}$. Using the same method but for $EB$ yields $EB=\frac{625}{48}$. As in previous solutions, by the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), $DM=\frac{5\sqrt{11}}{2}$. So, $\frac{DM}{EM}=\frac{24\sqrt{11}}{35}$.
Now, since we know both $CB$ and $EB$, we can find that $CE=\frac{527}{48}$. The height of $\Delta CME$ is the length from point $M$ to $CB$. Since $M$ is the midpoint of $AB$, the height is just $\frac{1}{2}\cdot7=\frac{7}{2}$. Using this, we can find that the area of $\Delta CMD$ is $\frac{1}{2}\cdot\frac{7}{2}\cdot\frac{527}{48}\cdot\frac{24\sqrt{11}}{35}=\frac{527\sqrt{11}}{40}$, giving our answer of $\boxed{578}$.
Solution by someonenumber011.