Difference between revisions of "1975 USAMO Problems/Problem 4"
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| − | Solution posted at | + | Solution with graph posted at |
http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1975Problem4 | 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 | by Vo Duc Dien | ||
Revision as of 13:20, 28 January 2010
Problem
Two given circles intersect in two points 
and 
. Show how to construct a segment 
passing through and terminating on the two circles such that 
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]](http://latex.artofproblemsolving.com/c/1/c/c1ce46c52cfe092b52af621b5ca152165e1e7826.png)
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 (Problems • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
