11. The lengths of the three altitudes of a triangle are in the ratio $3: 4: 6$. If the radius of the inscribed circle of the triangle is 1 , find the radius of the circumcircle of the triangle.
某三角形三條高的長度之比為 $3: 4: 6$ 。若該三角形內切圓的半徑是 1 , 求它的外接圓半徑。
11. $\frac{16}{5}$
11. Since the lengths of the altitudes are inversely proportional to the side lengths, the sides of the triangle are in ratio $\frac{1}{3}: \frac{1}{4}: \frac{1}{6}=4: 3: 2$. Let the side lengths be $4 k, 3 k, 2 k$. Then the triangle has perimeter $9 k$ and Heron's formula asserts that it has area
$$
k^{2} \sqrt{4.5(4.5-4)(4.5-3)(4.5-2)}=\frac{3 \sqrt{15}}{4} k^{2} .
$$

Since the in-radius of a triangle is equal to twice the area divided by the perimeter, we have
$$
2 \times \frac{3 \sqrt{15}}{4} k^{2} \div 9 k=1,
$$
which gives $k=\frac{6}{\sqrt{15}}$. Let $\theta$ be the angle opposite the shortest side and $R$ be the radius of the circumcircle. Then the sine law and cosine law give
$$
2 R=\frac{2 k}{\sin \theta}=\frac{2 k}{\sqrt{1-\cos ^{2} \theta}}=\frac{2 k}{\sqrt{1-\left(\frac{3^{2}+4^{2}-2^{2}}{2 \cdot 3 \cdot 4}\right)^{2}}}
$$
and hence $R=\frac{16}{5}$.