1975 USAMO Problems/Problem 4

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Problem

Two given circles intersect in two points $P$ and $Q$. Show how to construct a segment $AB$ passing through $P$ and terminating on the two circles such that $AP\cdot PB$ is a maximum.

[asy] size(150); defaultpen(fontsize(7)); pair A=(0,0), B=(10,0), P=(4,0), Q=(3.7,-2.5); draw(A--B); draw(circumcircle(A,P,Q)); draw(circumcircle(B,P,Q)); label("A",A,(-1,1));label("P",P,(0,1.5));label("B",B,(1,1));label("Q",Q,(-0.5,-1.5)); [/asy]

Solution with graph posted at

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1975Problem4

and here:

Let E and F be the centers of the small and big circles, respectively, and r and R be their respective radii.

Let M and N be the feet of E and F to AB, and α = ∠APE and ε = ∠BPF

We have:

AP × PB = 2r cosα × 2R cosε = 4 rR cosα cosε

AP × PB is maximum when the product cosα cosε is a maximum.

We have cosα cosε = ½ [cos(α + ε) + cos(α - ε)]

But α + ε = 180° - ∠EPF and is fixed, so is cos(α + ε)

So its maximum depends on cos(α - ε) which occurs when α = ε. To draw the line AB:

Draw a circle with center P and radius PE to cut the radius PF at H. Draw the line parallel to EH passing through P. This line meets the small and big circles at A and B, respectively.

by Vo Duc Dien

See also

1975 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5
All USAMO Problems and Solutions