Isosceles $\triangle{ABC}$ has a right angle at $C$. Point $P$ is inside $\triangle{ABC}$, such that $PA=11$, $PB=7$, and $PC=6$. Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$, where $a$ and $b$ are positive integers. What is $a+b$?
Given that $\triangle ABC$ is an isosceles right triangle, we can conclude that $AC=BC$. Let $AC=BC=s$.
Since $\triangle ABC$ is a right triangle, we can use the Pythagorean theorem:
\[AC^2 + BC^2 = AB^2\]
\[s^2 + s^2 = AB^2\]
\[2s^2=AB^2\]
\[AB=\sqrt{2} s\]
Now, let's draw the diagram and label the points:
[asy]
pair A,B,C,P;
A = (0,0);
B = (0,1);
C = (1,0);
P = (0.7,0.3);
draw(A--B--C--A);
draw(A--P);
draw(B--P);
draw(C--P);
label("$A$",A,SW);
label("$B$",B,NW);
label("$C$",C,SE);
label("$P$",P,N);
label("$s$",(A--C),S);
label("$s$",(B--C),NE);
label("$11$",(A--P),NW);
label("$7$",(B--P),NE);
label("$6$",(C--P),NW);
[/asy]
Using the law of cosines in $\triangle PAB$:
\[AB^2 = PA^2 + PB^2 - 2 \cdot PA \cdot PB \cdot \cos(\angle APB)\]
\[(\sqrt{2}s)^2 = 11^2 + 7^2 - 2 \cdot 11 \cdot 7 \cdot \cos(\angle APB)\]
\[2s^2 = 121 + 49 - 154 \cos(\angle APB)\]
We also notice that $\angle APC$ and $\angle BPC$ are both right angles. Since $\triangle ABC$ is an isosceles triangle with right angle at the vertex $C$, $\triangle APC$ and $\triangle BPC$ are congruent right triangles.
Therefore, we have:
\[\cos(\angle APB) = \cos(\angle APC + \angle BPC) = \cos(\angle APC) \cdot \cos(\angle BPC) - \sin(\angle APC) \cdot \sin(\angle BPC)\]
Since $\angle APC$ and $\angle BPC$ are right angles, we have $\cos(\angle APC) = 0$ and $\sin(\angle APC) = 1$, and similarly for $\angle BPC$.
Substituting these values into the equation, we get:
\[2s^2 = 121 + 49 - 154 \cdot 0 \cdot 1 = 170\]
Therefore, $s^2=85$, and $s=\sqrt{85}$.
Since $s=\sqrt{a+b\sqrt{2}}$, we have $a=85$ and $b=0$, so $a+b=85+0=\boxed{85}$.
To find the lengths of legs $\overline{AC}$ and $\overline{BC}$, we can use the Pythagorean Theorem. Since $\triangle{ABC}$ is a right triangle with a right angle at $C$, we have:
\begin{align*}
AC^2 + BC^2 &= AB^2
\end{align*}
We know that $PA=11$, $PB=7$, and $PC=6$. Let's consider $\triangle{APC}$ and $\triangle{BPC}$. We can use the Pythagorean Theorem in each triangle to find the lengths of $\overline{AC}$ and $\overline{BC}$.
In $\triangle{APC}$:
\begin{align*}
AC^2 + 6^2 &= 11^2 \\
AC^2 + 36 &= 121 \\
AC^2 &= 85
\end{align*}
In $\triangle{BPC}$:
\begin{align*}
BC^2 + 6^2 &= 7^2 \\
BC^2 + 36 &= 49 \\
BC^2 &= 13
\end{align*}
Now, we can substitute the values of $AC^2$ and $BC^2$ into the Pythagorean Theorem equation for $\triangle{ABC}$:
\begin{align*}
85 + 13 &= AB^2 \\
98 &= AB^2
\end{align*}
Since $AB$ is the hypotenuse of the right triangle $\triangle{ABC}$, we can conclude that $\overline{AB}=\sqrt{98}=\sqrt{2 \cdot 49}=7 \sqrt{2}$.
Therefore, $s = \sqrt{a + b \sqrt{2}} = \sqrt{98} = 7\sqrt{2}$.
Thus, $a + b = 98 + 0 = 98$, so the answer is $\boxed{98}$.
try this:
https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_23