1965 IMO Problems/Problem 5
Contents
Problem
Consider with acute angle . Through a point perpendiculars are drawn to and , the feet of which are and respectively. The point of intersection of the altitudes of is . What is the locus of if is permitted to range over (a) the side , (b) the interior of ?
Solution
Let . Equation of the line . Point . Easy, point . Point , . Equation of , equation of . Solving: . Equation of the first altitude: . Equation of the second altitude: . Eliminating from (1) and (2): a line segment . Second question: the locus consists in the .
Solution 2
This solution is a simplified version of the previous solution, and it provides more information. The idea is to just follow the degrees of the expressions and equations in involved. If we manage to conclude that the equation for is an equation of degree , then we will know that it is a line. We don't need to know the equation explicitly.
Just like in the previous solution, we use analytic (coordinate) geometry, but we don't care how the axes are chosen.
The coordinates of are expressions of degree in .
The equation for is an equation of degree in whose constant term is an expression of degree in .
TO BE CONTINUED. SAVING MID WAY, SO I DON'T LOSE WORK DONE SO FAR.
See Also
1965 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |