AIME 1987 · 第 9 题

Solution

Let $PC = x$. Since $\angle APB = \angle BPC = \angle CPA$, each of them is equal to $120^\circ$. By the Law of Cosines applied to triangles $\triangle APB$, $\triangle BPC$ and $\triangle CPA$ at their respective angles $P$, remembering that $\cos 120^\circ = -\frac12$, we have

$$AB^2 = 36 + 100 + 60 = 196, BC^2 = 36 + x^2 + 6x, CA^2 = 100 + x^2 + 10x$$ Then by the Pythagorean Theorem, $AB^2 + BC^2 = CA^2$, so

$$x^2 + 10x + 100 = x^2 + 6x + 36 + 196$$ and

$$4x = 132 \Longrightarrow x = \boxed{033}.$$

Note

This is the Fermat point of the triangle.

Video Solution by Pi Academy

https://youtu.be/fZAChuJDlSw?si=wJUPmgVRlYwazauh

~ smartschoolboy9