Three points O(0,0), P(a,a^2), Q(-b,b^2), a0, b0, lie on the parabola y=x^2.

Question

Three points $O(0,0),\ P(a,a^2),\ Q(-b,b^2)$, $a>0,\ b>0$, lie on the parabola $y=x^2$. Let $S_1$ be the area of the region bounded by the line $PQ$ and the parabola, and $S_2$ be the area of the triangle $OPQ$. If the minimum value of $S_1/S_2$ is $m/n$, $\gcd(m,n)=1$, then $m+n$ is:

Answer: 7

Accepted Answer

The parabolic segment area is $S_1=\dfrac{(a+b)^3}{6}$ (standard result), and

$$S_2=\frac12|x_Py_Q-x_Qy_P|=\frac12(ab^2+a^2b)=\frac{ab(a+b)}{2}.$$

$$\frac{S_1}{S_2}=\frac{(a+b)^2}{3ab}\ge\frac{4ab}{3ab}=\frac43\quad(\text{by AM--GM, equality at }a=b).$$

$$m=4,\ n=3,\qquad m+n=7.$$

Solutions · 1