Smallest Area Possible

Geometry Level 5

For a rectangle $ABCD$ with coordinates $A = (0,0), B = (5,0)$ , $C= (5,3) , D = (0,3)$ , let $P$ denote a varable point lying between the rectangle $ABCD$ .

And we define $d(P,L)$ as the perpendicular distance of point $P$ from line $L$ . Suppose

$d(P,AB) \leq \min\left[d(P,BC), d(P,CD), d(P,AD) \right].$

Find the area of the region in which $P$ lies.

Not original question. I like this and thought of sharing it with you guys.

The answer is 5.25.

2 solutions

Alan Guo
Nov 24, 2015

The definition of the P can be rephrased as $d(P,AB) = min[d(P,AB),d(P,AB),d(P,AB),d(P,AB)]$

Therefore, the boundary of $P$ is described by the intersection of pairwise perpendicular bisectors of $AB$ and the other sides.

Let $P(x,y) | 0\leq x\leq 5, 0\leq y \leq 3$ .

For $d(P,AB)\leq d(P,CD)$ , $y\leq 2$ .

For $d(P,AB)\leq d(P,AD)$ , $y\leq x$ .

For $d(P,AB)\leq d(P,BC)$ , $y\leq -x+5$ .

Thus, the locus is a trapezium, with parallel lengths of 2 and 5, height of 1.5, with area 5.25.

Lu Chee Ket
Dec 4, 2015

A 0 0
B 5 0
C 5 3
D 0 3
P x  y

AB  y
BC  5 - x
CD  3 - y
DA  x

Area = $\frac12 ((3.5 - 1.5) + (5 - 0)) 1.5 = 5.25$

Answer: $\boxed{5.25}$

Smallest Area Possible

Not original question. I like this and thought of sharing it with you guys.

The answer is 5.25.

2 solutions

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