# Concentric circles(IMO 1988/Problem 1)

Let us consider 2 concentric circle radii R and r (R > r) with centre O.We fix P on the small circle and consider the variable chord PA of the small circle. Points B and C lie on the large circle; B,P,C are collinear and BC is perpendicular to AP.

i.) For which values of $\angle OPA$ do you think is the sum $AB^2+BC^2+CA^2$ extremal?

ii.) What are the possible positions of the midpoints U of BA and V of AC as varies?

Source; IMO 1988/Problem 1.

I was thinking about some solution involving coordinate geometry but I am interested in other synthetic geometry solutions as well.(In fact, am I right when I believe the first question is asking for the possible values of the given sum?) Thank you!

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No, question 1 is asking for the values the angle OPA, such that the sum is maximum or minimum. – Ted Oct 21 '11 at 7:05
Thank you.In fact,I tried connecting A, B and C to O.With O(0,0) as the origin, I assigned coordinates to A, B and C and obtained relations of the type $p^2+q^2=R^2$ and so on.I tried using the Pythagoras theorem and used the previous results of the type I mentioned,but I always seemed to get stuck while trying to obtain the sum mentioned in the problem as expression independent of the co-ordinates I had specified.Your insight is appreciated! – Eisen Oct 21 '11 at 13:13
@SabyasachiMukherjee Are you still looking for solution, I can write one up for you. – Kirthi Raman Apr 12 '12 at 18:08

let $OH//AP,\angle APO=\theta$,we have : $OH=r\cos{\theta},AP=2OH=2r\cos{\theta},PH=r\sin{\theta},BH=\sqrt{BO^2-OH^2}=\sqrt{R^2-(r\cos{\theta})^2},BP=BH-PH=\sqrt{R^2-(r\cos{\theta})^2}-r\sin{\theta}, CP=BH+PH=\sqrt{R^2-(r\cos{\theta})^2}+r\sin{\theta},BC=2BH$

$AB^2=AP^2+BP^2=4r^2 \cos ^2{\theta}+R^2-r^2\cos ^2{\theta}+r^2\sin ^2{\theta}-2r\sin{\theta}\sqrt{R^2-(r\cos{\theta})^2}$

$AC^2=AP^2+CP^2=4r^2 \cos ^2{\theta}+R^2-r^2\cos ^2{\theta}+r^2\sin ^2{\theta}+2r\sin{\theta}\sqrt{R^2-(r\cos{\theta})^2}$

$AB^2+AC^2+BC^2=2R^2+6r^2 \cos ^2{\theta}+2r^2\sin ^2{\theta}+4R^2-4r^2 \cos ^2{\theta}=6R^2+2r^2$

so it is fixed value.

let $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3), P(-r,0) \to U \left( x_{u}=\dfrac{x_1+x_2}{2},y_{u}=\dfrac{y_1+y_2}{2}\right), V \left(x_{v}=\dfrac{x_1+x_3}{2},y_{v}=\dfrac{y_1+y_3}{2}\right)$

$k_{BP}=\dfrac{y_2}{x_2+r}=k_{CP}=\dfrac{y_3}{x_3+r}=-\dfrac{1}{k_{AP}}=-\dfrac{x_1+r}{y_1} \to$

$y_2y_1+x_2x_1+r^2+(x_1+x_2)r=0,$ and $y_3y_1+x_3x_1+r^2+(x_1+x_3)r=0$

$x_1^2+y_1^2+x_2^2+y_2^2=r^2+R^2 \to x_1^2+2x_1x_2+x_2^2+y_1^2+2y_1y_2+y_2^2+2r^2+2r(x_1+x_2)=R^2+r^2 \to (x_1+x_2)^2+2r(x_1+x_2)+r^2+(y_1+y_2)^2=R^2 \to 4x_u^2+4x_ur+r^2+4y_u^2=R^2 \to \left(x_u+\dfrac{r}{2}\right)^2+y_u^2=\left(\dfrac{R}{2}\right)^2$. with same method, $\left(x_v+\dfrac{r}{2}\right)^2+y_v^2=\left(\dfrac{R}{2}\right)^2$ , U and V on a circle, each take half circle.

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