2000 AIME II 第 10 题

The center lies on all four angle bisectors, so the half-angle at $A$ satisfies $\tan\frac{A}{2} = \frac{r}{19},$ and similarly at the other vertices

The four half-angles sum to $180^\circ,$ so $\tan(\alpha + \gamma) = -\tan(\beta + \delta);$ the addition formula gives a linear equation in $r^2$

Let the incircle have center $I$ and radius $r.$ The tangent lengths from $A,$ $B,$ $C,$ $D$ are $19,$ $26,$ $37,$ $23,$ and $I$ lies on each angle bisector, so the half-angles $\alpha, \beta, \gamma, \delta$ at the four vertices satisfy $\tan\alpha = \frac{r}{19},$ $\tan\beta = \frac{r}{26},$ $\tan\gamma = \frac{r}{37},$ $\tan\delta = \frac{r}{23},$ with $\alpha + \beta + \gamma + \delta = 180^\circ.$

Then $\tan(\alpha + \gamma) = -\tan(\beta + \delta),$ and the tangent addition formula turns this into $$\begin{aligned} &\frac{\frac{r}{19} + \frac{r}{37}}{1 - \frac{r^2}{19 \cdot 37}} \\ &= -\frac{\frac{r}{26} + \frac{r}{23}}{1 - \frac{r^2}{26 \cdot 23}}, \end{aligned}$$ i.e. $$\frac{56r}{703 - r^2} = \frac{49r}{r^2 - 598}.$$

Cross-multiplying gives $56r^2 - 56 \cdot 598$ $= 49 \cdot 703 - 49r^2,$ so $105r^2 = 33488 + 34447 = 67935$ and $r^2 = 647.$