Two given circles intersect in two points $ P$ and $ Q$. Show how to construct a segment $ AB$ passing through $ P$ and terminating on the circles such that $ AP \cdot PB$ is a maximum.
Problem
Source: 1975 USAMO Problem 4
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modularmarc101
15.03.2010 16:05
By AM-GM, the maximum of $ AP \cdot PB$ is $ \frac{AB^2}{4}$, and is attained when $ AP = PB$. I just don't know how to construct that segment..
Kenny O
15.03.2010 16:40
@ modularmarc101 , I think that In this case AM-GM doesnt work. I do believe that the maximum AP.PB is attained when AB is parallel to the line that joins the centers of both circles, and this does not mean that AP=PB for different circles. am I right?
bpgbcg
15.03.2010 22:11
It is well known that considering all such A and B, $ \angle{AQB}$ is constant. Let $ r_1$ be the radius of the circle which A is on and $ r_2$ be the radius of the circle which B is on. Then $ AP\cdot{PB}=(2r_1sin(\angle{AQP}))(2r_2sin(\angle{BQP}))=4r_1r_2sin(\angle{AQP})sin(\angle{BQP})$, so we just must maximize $ sin(\angle{AQP})sin(\angle{BQP})$. But this equals $ \frac{1}{2}(cos(\angle{AQP}-\angle{BQP})-cos(\angle{AQB}))$, which is maximized when the two angles are equal. Thus the maximum occurs when $ PQ$ is the angle bisector. To construct this, construct the two external tangents and let them meet at X. Let XP meet the circles at A and B. The two arcs AP and PB are the same (with respect to their circles), since they are homothetic. Thus the angles are equal, so we are done.
SnowEverywhere
17.03.2010 05:18
@modularmarc101: Since $ AB$ is not constant for all $ A$ and $ B$, this is not necessarily the maximal condition.
Vo Duc Dien
23.11.2010 23:09
http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1975Problem4
huashiliao2020
14.04.2023 03:23
bpgbcg wrote:
It is well known that considering all such A and B, $ \angle{AQB}$ is constant. Let $ r_1$ be the radius of the circle which A is on and $ r_2$ be the radius of the circle which B is on. Then $ AP\cdot{PB}=(2r_1sin(\angle{AQP}))(2r_2sin(\angle{BQP}))=4r_1r_2sin(\angle{AQP})sin(\angle{BQP})$, so we just must maximize $ sin(\angle{AQP})sin(\angle{BQP})$. But this equals $ \frac{1}{2}(cos(\angle{AQP}-\angle{BQP})-cos(\angle{AQB}))$, which is maximized when the two angles are equal. Thus the maximum occurs when $ PQ$ is the angle bisector. To construct this, construct the two external tangents and let them meet at X. Let XP meet the circles at A and B. The two arcs AP and PB are the same (with respect to their circles), since they are homothetic. Thus the angles are equal, so we are done.
How did you get AP*PB=2r_1sinAQP*2r_2sinBQP? I got that <AQB is constant since it is 180-PAQ-PBQ, which are always subtending arc PQ each. hints to solve it different way is also accepted
S.Das93
14.04.2023 04:16
We claim that the maximum value of $AP \cdot PB$ is attained when $A, P, B$ are collinear on a line parallel to $O_1O_2$, the line joining the centers of the circles. Since $PQ$ is the radical axis of circles $O_1$ and $O_2$, we must also have that $Pow_A=Pow_B=Pow_P$, which only occurs when $AB \perp PQ$. This happens when $AB || O_1O_2$. $\blacksquare$