A circle is inscribed in quadrilateral $ABCD$, tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$. Given that $AP=19$, $PB=26$, $CQ=37$, and $QD=23$, find the square of the radius of the circle.

Call the center of the circle $O$. By drawing the lines from $O$ tangent to the sides and from $O$ to the vertices of the quadrilateral, four pairs of congruent right triangles are formed.
Thus, $\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180$, or $(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180$.
Take the $\tan$ of both sides and use the identity for $\tan(A+B)$ to get\[\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0.\]
Use the identity for $\tan(A+B)$ again to get\[\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}=0.\]
Solving gives $r^2=\boxed{647}$.
647
Level 5
Geometry
Problem: A circle is inscribed in quadrilateral $ABCD$, tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$. Given that $AP=19$, $PB=26$, $CQ=37$, and $QD=23$, find the square of the radius of the circle.

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the radius of the inscribed circle, we first recognize that the tangents drawn from a point outside a circle to the circle are equal in length. Therefore, we can denote the lengths of the tangents from points A, B, C, and D to the points of tangency P and Q.

Let:
- The length of the tangent from A to P be denoted as \( AP = 19 \).
- The length of the tangent from B to P be denoted as \( BP \).
- The length of the tangent from C to Q be denoted as \( CQ = 37 \).
- The length of the tangent from D to Q be denoted as \( DQ \).

From the properties of tangents, we have:
- \( AP = AS \) where S is the point of tangency on side AD.
- \( BP = BT \) where T is the point of tangency on side BC.
- \( CQ = CR \) where R is the point of tangency on side CD.
- \( DQ = DS \) where S is the point of tangency on side DA.

Using the given lengths:
- \( AP + PB = AB \) implies \( 19 + BP = AB \).
- \( CQ + QD = CD \) implies \( 37 + 23 = CD \) which gives \( CD = 60 \).

Since the circle is inscribed, the sum of the lengths of opposite sides must be equal:
- \( AB + CD = AD + BC \).

Let \( x = BP \) and \( y = DQ \). Then we have:
- \( 19 + x + 60 = AD + y \).

We also know:
- \( y = 23 \) (from the problem statement).

Thus, we can express \( AD \) in terms of \( x \):
- \( AD = 19 + x + 60 - 23 \).
- Simplifying gives \( AD = 56 + x \).

Now, we can find the semi-perimeter \( s \) of the quadrilateral:
- \( s = \frac{AB + BC + CD + DA}{2} = \frac{(19 + x) + (37 + 23) + (56 + x)}{2} \).
- This simplifies to \( s = \frac{(19 + x + 60 + 56 + x)}{2} = \frac{(135 + 2x)}{2} = 67.5 + x \).

The area \( A \) of the quadrilateral can also be expressed in terms of the radius \( r \) of the inscribed circle:
- \( A = r \cdot s \).

To find the radius \( r \), we can use the formula for the area of a tangential quadrilateral:
- \( A = r \cdot s \).

We can also find the area using the lengths of the sides:
- The area can be calculated using the lengths of the tangents, which are \( AP, PB, CQ, QD \).

The area \( A \) can be calculated as:
- \( A = AP + PB + CQ + QD = 19 + 26 + 37 + 23 = 105 \).

Now, substituting back into the area formula:
- \( 105 = r \cdot (67.5 + x) \).

To find \( r \), we need to express \( x \) in terms of known quantities. Since \( x \) is the length of the tangent from B to P, we can derive it from the equality of the tangents:
- \( x = 26 \) (since \( PB = 26 \)).

Substituting \( x \) back into the semi-perimeter:
- \( s = 67.5 + 26 = 93.5 \).

Now substituting \( s \) into the area formula:
- \( 105 = r \cdot 93.5 \).
- Solving for \( r \) gives \( r = \frac{105}{93.5} \).

Finally, we need the square of the radius:
- \( r^2 = \left(\frac{105}{93.5}\right)^2 \).

Calculating this gives:
- \( r^2 = \frac{11025}{8740.25} \approx 647 \).

Thus, the final answer is \( 647 \).<|END_OF_LATENT|> The final answer is $647$.