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

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==Solution== | ==Solution== | ||

− | + | <ol type="i"> | |

− | + | <li> | |

We claim that the value <math>BC^2+CA^2+AB^2</math> stays constant as <math>\angle OPA</math> varies, and thus achieves its maximum at all value of <math>\angle OPA</math>. We have from the Pythagorean Theorem that <math>CA^2=AP^2+PC^2</math> and <math>AB^2=AP^2+PB^2</math> and so our expression becomes | We claim that the value <math>BC^2+CA^2+AB^2</math> stays constant as <math>\angle OPA</math> varies, and thus achieves its maximum at all value of <math>\angle OPA</math>. We have from the Pythagorean Theorem that <math>CA^2=AP^2+PC^2</math> and <math>AB^2=AP^2+PB^2</math> and so our expression becomes | ||

− | < | + | <cmath> BC^2+PB^2+PC^2+2AP^2=2BC^2+2AP^2-2PB\cdot PC </cmath> |

− | Since <math>PB\cdot PC</math> is the power of the point <math>P</math>, it stays constant as <math>A</math> varies. Thus, we are left to prove that the value <math>BC^2+AP^2</math> stays constant as <math>\angle OPA</math> varies. Let <math>G</math> be the midpoint of <math>AP</math> and let <math>H</math> be the midpoint of <math>BC</math>. | + | Since <math>PB\cdot PC</math> is the power of the point <math>P</math>, it stays constant as <math>A</math> varies. Thus, we are left to prove that the value <math>BC^2+AP^2</math> stays constant as <math>\angle OPA</math> varies. Let <math>G</math> be the midpoint of <math>AP</math> and let <math>H</math> be the midpoint of <math>BC</math>. Since <math>OG</math> is perpendicular to <math>AP</math>, we find that <math>PG=r\cos OPA</math>. Similarly, we find that <math>OH=r\sin OPC=r\cos OPA</math>. Thus, by the Pythagorean Theorem, we have |

− | < | + | <cmath> |

− | + | \begin{align*} | |

− | + | BC^2& =4(R^2-r^2\cos^2 OPA) \\ | |

+ | AP^2&=4r^2\cos^2OPA | ||

+ | \end{align*} | ||

+ | </cmath> | ||

Now it is obvious that <math>BC^2+AP^2=4R^2</math> is constant for all values of <math>\angle OPA</math>. | Now it is obvious that <math>BC^2+AP^2=4R^2</math> is constant for all values of <math>\angle OPA</math>. | ||

+ | </li> | ||

+ | <li> | ||

− | + | We claim that all points <math>U,V</math> lie on a circle centered at the midpoint of <math>OP</math>, <math>M</math> with radius <math>\frac{R}{2}</math>. Let <math>T</math> be the midpoint of <math>UV</math>. Since <math>H</math> is the midpoint of <math>BC</math>, it is clear that the projection of <math>T</math> onto <math>BC</math> is the midpoint of <math>H</math> and <math>P</math> (the projection of <math>A</math> onto <math>BC</math>). Thus, we have that <math>MT</math> is perpendicular to <math>UV</math> and thus the triangle <math>MUV</math> is isosceles. We have | |

− | + | <cmath> | |

− | We claim that all points <math>U,V</math> lie on a circle centered at the midpoint of <math>OP</math>, <math>M</math> with radius <math>\frac{R}{2}</math>. Let <math>T</math> be the midpoint of <math>UV</math>. Since <math>H</math> is the midpoint of <math>BC</math>, it is clear that the projection of <math>T</math> onto <math>BC</math> is the midpoint of <math>H</math> and <math>P</math> (the projection of <math>A</math> onto <math>BC</math>). Thus, we have that <math>MT</math> is perpendicular to <math>UV</math> and thus the triangle <math>MUV</math> is | + | \begin{align*} |

+ | UT &=\frac{1}{2}UV=\frac{1}{4}BC \mbox{ and } \\ | ||

+ | MT &=\frac{1}{2}PG=\frac{1}{4}AP. | ||

+ | \end{align*} | ||

+ | </cmath> | ||

+ | Thus, from the Pythagorean Theorem we have | ||

+ | <cmath> MV^2=MU^2=\frac{1}{16}\left(BC^2+AP^2\right).</cmath> | ||

+ | Since we have shown already that <math>BC^2+AP^2=4R^2</math> is constant, we have that <math>MV=MU=\frac{R}{2}</math> and the locus of points <math>U,V</math> is indeed a circle of radius <math>\frac{R}{2}</math> with center <math>M</math>. | ||

+ | </li> | ||

+ | </ol> |

## Revision as of 20:27, 9 May 2011

## Problem

Consider 2 concentric circles with radii and () with center . Fix on the small circle and consider the variable chord of the small circle. Points and lie on the large circle; are collinear and is perpendicular to .

i.) For which values of is the sum extremal?

ii.) What are the possible positions of the midpoints of and of as varies?

## Solution

- We claim that the value stays constant as varies, and thus achieves its maximum at all value of . We have from the Pythagorean Theorem that and and so our expression becomes Since is the power of the point , it stays constant as varies. Thus, we are left to prove that the value stays constant as varies. Let be the midpoint of and let be the midpoint of . Since is perpendicular to , we find that . Similarly, we find that . Thus, by the Pythagorean Theorem, we have Now it is obvious that is constant for all values of .
- We claim that all points lie on a circle centered at the midpoint of , with radius . Let be the midpoint of . Since is the midpoint of , it is clear that the projection of onto is the midpoint of and (the projection of onto ). Thus, we have that is perpendicular to and thus the triangle is isosceles. We have Thus, from the Pythagorean Theorem we have Since we have shown already that is constant, we have that and the locus of points is indeed a circle of radius with center .