Answer to Question #150070 in Combinatorics | Number Theory for Wesam

Question

Let λ be a real number. Suppose that if ABCD is any convex cyclic quadrilateral such that AC = 4, BD = 5, and AB ⊥ CD, then the area of ABCD is at least λ. Then the greatest possible value of λ is m/n, where m and n are positive integers with gcd(m,n) = 1. Compute 100m+n.

Answer 1

The area of a quadrilateral with sides a, b, c, d:

$S=\frac{1}{2}(ac+bd)sin\theta$ , where $\theta$ is an angle between diagonals

By Ptolemy's Theorem:

$ac+bd=ef$ , where e, f are diagonals

In our case:

$e=AC, f=BD$

Then:

$S=\frac{4\cdot5}{2}sin\theta=10sin\theta$

$S_{max}=10sin90\degree=10=m/n$

$m=10, n=1$ since $gcd(m,n) = 1$

So:

$100m+n=100\cdot10+1=1001$