Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1988 IMO Problems/Problem 1"

(→‎Solution)
 
Line 41: Line 41:
 
</li>
 
</li>
 
</ol>
 
</ol>
 +
 +
 +
== See Also == {{IMO box|year=1988|before=First Question|num-a=2}}
 +
 +
[[Category:Olympiad Geometry Problems]]

Latest revision as of 10:32, 30 January 2021

Problem

Consider 2 concentric circles with radii $R$ and $r$ ($R>r$) with center $O$. Fix $P$ on the small circle and consider the variable chord $AP$ of the small circle. Points $B$ and $C$ lie on the large circle; $B,P,C$ are collinear and $BC$ is perpendicular to $AP$.

  1. For which values of $\angle OPA$ is the sum $BC^2+CA^2+AB^2$ extremal?
  2. What are the possible positions of the midpoints $U$ of $AB$ and $V$ of $AC$ as $A$ varies?

Solution

  1. We claim that the value $BC^2+CA^2+AB^2$ stays constant as $\angle OPA$ varies, and thus achieves its maximum at all value of $\angle OPA$. We have from the Pythagorean Theorem that $CA^2=AP^2+PC^2$ and $AB^2=AP^2+PB^2$ and so our expression becomes \[BC^2+PB^2+PC^2+2AP^2=2BC^2+2AP^2-2PB\cdot PC\] Since $PB\cdot PC$ is the power of the point $P$, it stays constant as $A$ varies. Thus, we are left to prove that the value $BC^2+AP^2$ stays constant as $\angle OPA$ varies. Let $G$ be the midpoint of $AP$ and let $H$ be the midpoint of $BC$. Since $OG$ is perpendicular to $AP$, we find that $PG=r\cos OPA$. Similarly, we find that $OH=r\sin OPC=r\cos OPA$. Thus, by the Pythagorean Theorem, we have \begin{align*} BC^2& =4(R^2-r^2\cos^2 OPA) \\ AP^2&=4r^2\cos^2OPA \end{align*} Now it is obvious that $BC^2+AP^2=4R^2$ is constant for all values of $\angle OPA$.
  2. We claim that all points $U,V$ lie on a circle centered at the midpoint of $OP$, $M$ with radius $\frac{R}{2}$. Let $T$ be the midpoint of $UV$. Since $H$ is the midpoint of $BC$, it is clear that the projection of $T$ onto $BC$ is the midpoint of $H$ and $P$ (the projection of $A$ onto $BC$). Thus, we have that $MT$ is perpendicular to $UV$ and thus the triangle $MUV$ is isosceles. We have \begin{align*} UT &=\frac{1}{2}UV=\frac{1}{4}BC \mbox{ and } \\ MT &=\frac{1}{2}PG=\frac{1}{4}AP. \end{align*} Thus, from the Pythagorean Theorem we have \[MV^2=MU^2=\frac{1}{16}\left(BC^2+AP^2\right).\] Since we have shown already that $BC^2+AP^2=4R^2$ is constant, we have that $MV=MU=\frac{R}{2}$ and the locus of points $U,V$ is indeed a circle of radius $\frac{R}{2}$ with center $M$.


See Also

1988 IMO (Problems) • Resources
Preceded by
First Question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1988_IMO_Problems/Problem_1&oldid=144054"