Question #150070
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.
1
Expert's answer
2020-12-16T20:26:45-0500

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

S=12(ac+bd)sinθS=\frac{1}{2}(ac+bd)sin\theta , where θ\theta is an angle between diagonals

By Ptolemy's Theorem:

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

In our case:

e=AC,f=BDe=AC, f=BD

Then:

S=452sinθ=10sinθS=\frac{4\cdot5}{2}sin\theta=10sin\theta

Smax=10sin90°=10=m/nS_{max}=10sin90\degree=10=m/n

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

So:

100m+n=10010+1=1001100m+n=100\cdot10+1=1001


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