China Girls' Mathematical Olympiad

Question

Let A be the closed domain on the plane delimited by three lines x = 1, y = 0, and y = t(2x - t), where 0 < t < 1. Prove that the surface of any triangle inside the domain A with P(t, t^2) and Q(1, 0) as two of its vertices cannot exceed 14. FIGURE_0

Accepted Answer

It is easy to observe that the domain is a closed triangle. Its three vertices are

B (\frac{t}{2}, 0)

,

Q (1, 0)

and

C (1, t (2 - t))

. Pick a point

X

inside the

△ B QC

, then the area of

△ P QX

is equal to half of the product of

P Q

with the distance from

X

to

P Q

. So the area of

P QX

takes its maximum value when the distance from

X

to

P Q

is maximized, i.e., when

X

coincides with

B

or

C

. The area of

△ P QB

is

\frac{1}{2} (1 - \frac{t}{2}) t^{2} = \frac{1}{4} (2 - t) t^{2} \leq \frac{1}{4} (2 - t) t \leq \frac{1}{4} (\frac{2 - t + t}{2})^{2} = \frac{1}{4};

the area of

△ P QC

is

\frac{1}{2} (1 - t) (2 t - t^{2}) = \frac{1}{4} 2 t (1 - t) (2 - t) \leq \frac{1}{4} (\frac{2 t + 1 - t + 2 - t}{3})^{3} = \frac{1}{4} .

Hence, in the domain

A

, any triangle with

P, Q

as two of its vertices cannot have an area that exceeds

\frac{1}{4}

.

□