Difference between revisions of "1979 IMO Problems/Problem 4"
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Hence, the point <math>R</math> for which the ratio <math>\frac{QP+PR}{QR}</math> is maximum is the point of intersection of the ray <math>PT</math> with the sphere with center <math>P</math> and radius <math>PQ</math> (or, respectively, it can be any arbitrary point on the intersection of the plane <math>p</math> with the sphere with center <math>P</math> and radius <math>PQ</math> if <math>PQ\perp p</math>). | Hence, the point <math>R</math> for which the ratio <math>\frac{QP+PR}{QR}</math> is maximum is the point of intersection of the ray <math>PT</math> with the sphere with center <math>P</math> and radius <math>PQ</math> (or, respectively, it can be any arbitrary point on the intersection of the plane <math>p</math> with the sphere with center <math>P</math> and radius <math>PQ</math> if <math>PQ\perp p</math>). | ||
− | == See Also == {{IMO box|year=1979|num-b= | + | This solution was posted and copyrighted by darij grinberg. The original thread for this problem can be found here: [https://aops.com/community/p392254] |
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+ | == See Also == {{IMO box|year=1979|num-b=3|num-a=5}} |
Latest revision as of 23:01, 29 January 2021
Problem
We consider a point in a plane and a point . Determine all the points from for whichis maximum.
Solution
Let be the orthogonal projection of the point on the plane . Then, the line is the orthogonal projection of the line on the plane , and thus forms the least angle with the line among all lines through the point which lie in the plane ; hence, , and equality holds if and only if the point lies on the ray (the only exception is when ; in this case, , so the ray is undefined, and equality holds for all points in the plane , since we always have ).
From , it follows that ; also, since the angles and are , their half-angles and are < 90°, so that from we can conclude that . Equality holds, as in the above, if and only if the point lies on the ray (and, respectively, for all points in the plane if ).
On the other hand, we obviously have with equality if and only if , i. e. if and only if < QRP = < PQR, i. e. if and only if triangle PQR is isosceles with base , i. e. if and only if , i. e. if and only if the point lies on the sphere with center and radius .
Now, applying the Mollweide theorem in triangle , we get
(since ) (since ),
and equality holds here if and only if equality holds in both of the inequalities and that we have used, i. e. if and only if the point lies both on the ray (this condition should be ignored if ) and on the sphere with center and radius .
Hence, the point for which the ratio is maximum is the point of intersection of the ray with the sphere with center and radius (or, respectively, it can be any arbitrary point on the intersection of the plane with the sphere with center and radius if ).
This solution was posted and copyrighted by darij grinberg. The original thread for this problem can be found here: [1]
See Also
1979 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |