B3. In the figure, $B C$ is a diameter of the circle, where $B C=\sqrt{901}, B D=1$, and $D A=16$. If $E C=x$, what is the value of $x$ ?
Solution: The answer is 26 .

Solution 1: Since $B C$ is the diameter of the circle, $\angle B D C=\angle B E C=90^{\circ}$. By the Pythagorean Theorem, we have
$$
C D=\sqrt{B C^{2}-B D^{2}}=\sqrt{901-1^{2}}=\sqrt{900}=30 .
$$

Since $\angle B D C=90^{\circ}, \angle A D C=90^{\circ}$. Then by the Pythagorean Theorem, we have
$$
A C=\sqrt{A D^{2}+D C^{2}}=\sqrt{16^{2}+30^{2}}=34 .
$$

Since $x=C E, A E=34-x$. We need to determine $x$. By the Pythagorean Theorem, we have $B E=\sqrt{B A^{2}-A E^{2}}=\sqrt{B C^{2}-C E^{2}}$. Hence,
$$
B A^{2}-A E^{2}=B C^{2}-C E^{2} \text {. }
$$

Note that $B A=B D+D A=16+1=17$. Therefore,
$$
\begin{array}{rlrl} 
& & 17^{2}-(34-x)^{2} & =901-x^{2} \\
\Rightarrow & x^{2}+289 & =(x-34)^{2}+901 \\
\Rightarrow & x^{2}+289 & =x^{2}-68 x+1156+901 \\
\Rightarrow & 68 x & =1768 .
\end{array}
$$

Therefore, $x=1768 / 68=(17 \times 104) /(17 \times 4)=104 / 4=26$. Hence, $E C=26$.

Solution 2: As in Solution 1, $\angle B D C=\angle B E C=90^{\circ}, C D=30$ and $A C=34$. By computing the area of $\triangle A B C$ in two different ways, we have
$$
\text { the area of } \triangle A B C=\frac{1}{2} \times A B \times D C=\frac{1}{2} \times A C \times B E \text {. }
$$

Therefore, $A B \cdot D C=A C \cdot B E$. Hence, $17 \cdot 30=34 \cdot B E$. Therefore, $30=2 \cdot B E$. Equivalently, $B E=15$. Therefore, by the Pythagorean Theorem,
$$
E C=\sqrt{B C^{2}-B E^{2}}=\sqrt{901-15^{2}}=\sqrt{676}=26 .
$$

Solution 3: As in Solution 1, $\angle B D C=\angle B E C=90^{\circ}, C D=30$ and $A C=34$. Compare $\triangle A D C$ and $\triangle A E B$. Note that $\angle A D C=\angle A E B=90^{\circ}$ and $\angle D A C=\angle E A B$. Therefore, $\triangle A D C$ is similar to $\triangle A E B$. Consequently,
$$
\frac{A D}{A E}=\frac{A C}{A B} .
$$

Therefore, $A D \cdot A B=A E \cdot A C$. Note that $A B=A D+B D=16+1=17$. Hence, $16 \times 17=A E \cdot 34$. Therefore, $A E=8$. We then conclude that $E C=A C-A E=34-8=26$.