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(IMO 1988/1) Consider two circles of radii $R$ and $r$ $(R > r)$ with the same center. Let $P$ be a fixed point on the smaller circle and $B$ a variable point on the larger circle. The line $BP$ meets the larger circle again at $C$. The perpendicular $l$ to $BP$ at $P$ meets the smaller circle again at $A$. (As per our convention, if $l$ is tangent to the circle at $P,$ then we take $A = P$.)

(i) Find the set of values of $BC^2$ + $CA^2$ + $AB^2$

(ii) Find the locus of the midpoint of $AB$.

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See that website: artofproblemsolving.com/Forum/… –  user31029 Jun 19 '12 at 4:58
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2 Answers 2

$X$ is midpoint of $AB$. Easy to see that $XC$ divides $OP$ in $1:2$ ratio. So a $-0.5$ homothety centered at midpoint of $OP$ sends $C$ to $X$. Since $C$ moves on a circle so $X$ moves on a circle too and the first part is obvious.

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Adding a few more perpendiculars, we augment $PA$ to a rectangle $\square PAQA^\prime$ with diagonal $PQ$ a diameter of the smaller circle. (We've exploited the fact that any angle inscribed in a semicircle is a right angle.) This makes $\square AQBC$ an isosceles trapezoid, whose height we'll denote $q$, smaller base $p$, and larger base $p+2s$.

enter image description here

Note that Pythagoras gives us

$$|PQ|^2 = p^2 + q^2 = 4 r^2 \qquad |CA|^2 = q^2 + s^2 \qquad |AB|^2 = q^2 + (p+s)^2$$

Moreover, the (unsigned) Power of Point $P$ relative to the big circle is $$R^2 - r^2 = |PC| |PB| = s \left( p + s \right)$$

Therefore, $$\begin{align} |AB|^2 + |BC|^2 + |CA|^2 &= \left( q^2 + \left( p + s \right)^2 \right) + \left( p + 2 s \right)^2 + \left( q^2 + s^2 \right) \\ &= 2 \left( p^2 + q^2 \right) + 6 s \left( p + s \right) \\ &= 8 r^2 + 6 \left( R^2 - r^2 \right) \\ &= 2 \left( 3 R^2 + r^2 \right) \end{align}$$

(Similar algebra ---and an identical result--- arises from applying Ptolemy's theorem to the (necessarily-cyclic) isosceles trapezoid: $$|CA| |BQ| + |AQ| |BC| = |AB||QC|$$ where $|BQ| = |CA|$ and $|QC| = |AB|$.)

Consequently, the sum-of-squares is independent of the location of point $B$, so that the answer to (i) is the singleton set containing $2\left( 3R^2 + r^2\right)$.

For (ii), extend the trapezoid's shorter base to match the longer, obtaining rectangle $\square BCC^\prime B^\prime$.

enter image description here

The midpoint, $M$, of $AB$ is always the midpoint of $PB^\prime$, whereas $B^\prime$ is always a point on the larger circle. Thus, $M$ is a dilation of $B^\prime$ in $P$ with scale factor $1/2$, and the locus of $M$ is the corresponding dilation of the locus of $B^\prime$ (aka, the big circle).

Note: The locus of $N$, the midpoint of $AC$ (and of $PC^\prime$), is the same circle. Likewise with the midpoints of $PC$ and $PB$.

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