Difference between revisions of "1965 IMO Problems/Problem 5"
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This solution is elementary, it does not use analytic geometry. | This solution is elementary, it does not use analytic geometry. | ||
− | + | Let <math>A_1</math> be the foot of the perpendicular from <math>A</math> to <math>OB</math>, and | |
+ | <math>B_1</math> be the foot of the perpendicular from <math>B</math> to <math>OA</math>. Construct | ||
+ | <math>H</math> as described in the statement of the problem. | ||
+ | [[File:prob_1965_5.png|900px]] | ||
+ | We will prove that <math>H \in A_1B_1</math>. | ||
+ | First, as shown in the second picture, take <math>MQ \perp OB</math>, and let <math>H_1</math> | ||
+ | be the intersection of <math>A_1B_1</math> with the perpendicular from <math>Q</math> to <math>OA</math>. | ||
+ | From the triangle <math>\triangle BAA_1</math> we have | ||
+ | <math>\frac{AM}{MB} = \frac{A_1Q}{QB}</math> because <math>AA_1 \parallel MQ</math>. From the | ||
+ | triangle <math>\triangle A_1BB_1</math> we have | ||
+ | <math>\frac{A_1Q}{QB} = \frac{A_1H_1}{H_1B_1}</math> because <math>QH_1 \parallel BB_1</math>. | ||
+ | So, <math>\frac{AM}{MB} = \frac{A_1H_1}{H_1B_1}</math>. | ||
+ | Second, as shown in the third picture, take <math>MP \perp OA</math>, and let <math>H_2</math> | ||
+ | be the intersection of <math>A_1B_1</math> with the perpendicular from <math>P</math> to <math>OB</math>. | ||
+ | By a similar argument, we have <math>\frac{AM}{MB} = \frac{A_1H_2}{H_2B_1}</math>. | ||
− | + | It follows that <math>H_1 = H_2</math> since the two points divide <math>A_1B_1</math> in the | |
+ | same ratio. This is the point <math>H</math> from the statement of the problem. | ||
+ | |||
+ | The solution to the problem is now completed by repeating the last two | ||
+ | paragraphs from Solution 2. | ||
+ | |||
+ | (Solution by pf02, October 2024) | ||
== See Also == | == See Also == | ||
{{IMO box|year=1965|num-b=4|num-a=6}} | {{IMO box|year=1965|num-b=4|num-a=6}} |
Latest revision as of 19:31, 31 October 2024
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, it fills in some gaps. and it provides more information.
Just like in the previous solution, we use analytic (coordinate) geometry, but we don't care how the axes are chosen.
Let have coordinates , and let with . The idea is to just follow the degrees of the expressions and equations in involved as we make the computations for obtaining the coordinates of , and the equation of the curve is on. We will see that the equation for is an equation of degree , so we will know that it is a line. We don't need to write out the equation explicitly.
The coordinates of are expressions of degree in .
The equation for (the perpendicular from to ) is an equation of degree in with constant coefficients for , and whose constant term is an expression of degree in .
The coordinates of (the foot of the perpendicular from to ) are expressions of degree in .
The equation of the perpendicular from to is of degree in , with constant coefficients for , and whose constant term is an expression of degree in . This corresponds to equation (2) in the above solution.
Similarly, the equation of the perpendicular from to is of degree in , with constant coefficients for , and whose constant term is an expression of degree in . This corresponds to equation (1) in the above solution.
Now, in principle, we would have to solve the system of two equations (1) and (2) to obtain the coordinates of as expressions of , and then eliminate to obtain the equation in for . As a shortcut, we can eliminate directly from the two equations (1) and (2). Either way, the result is an equation of degree in .
This tells us that the locus is on a line. We just need to specify which set of points on this line is the locus. And, we want to make the line explicit.
The previous solution, with a good amount of hand waving, tells us that the solution is "a line segment ". (On top of the hand waving the solution uses the unhappy notation for and for , which is bad because has already been used!) We will do better than that.
Let be the foot of the perpendicular from to , and be the foot of the perpendicular from to . (For this paragraph see the picture shown in Solution 3.) Consider the limit situation when . Then , and . It follows that the intersection of the perpendiculars from to and to is . Similarly, the limit situation when yields . Now it is reasonable to say that when moves from to , moves from to . So, the locus is the line segment joining the feet of the perpendiculars in from . This answers question (a).
For part (b) of the problem, with a good amount of hand waving, the previous solution says "the locus consists in the ". We justify this by pointing out that if is inside , then we can take the triangle , such that , , going through and parallel to . Then will be on the corresponding segment determined by the feet of the perpendiculars in . Conversely, it is easy to see that any point is on a segment obtained from a triangle , and is obtained from a point . This answers question (b).
(Solution by pf02, October 2024)
Solution 3
This solution is elementary, it does not use analytic geometry.
Let be the foot of the perpendicular from to , and be the foot of the perpendicular from to . Construct as described in the statement of the problem.
We will prove that .
First, as shown in the second picture, take , and let be the intersection of with the perpendicular from to . From the triangle we have because . From the triangle we have because . So, .
Second, as shown in the third picture, take , and let be the intersection of with the perpendicular from to . By a similar argument, we have .
It follows that since the two points divide in the same ratio. This is the point from the statement of the problem.
The solution to the problem is now completed by repeating the last two paragraphs from Solution 2.
(Solution by pf02, October 2024)
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 |