Difference between revisions of "1965 IMO Problems/Problem 5"
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== Problem == | == Problem == | ||
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Consider <math>\triangle OAB</math> with acute angle <math>AOB</math>. Through a point <math>M \neq O</math> perpendiculars are drawn to <math>OA</math> and <math>OB</math>, the feet of which are <math>P</math> and <math>Q</math> respectively. The point of intersection of the altitudes of <math>\triangle OPQ</math> is <math>H</math>. What is the locus of <math>H</math> if <math>M</math> is permitted to range over (a) the side <math>AB</math>, (b) the interior of <math>\triangle OAB</math>? | Consider <math>\triangle OAB</math> with acute angle <math>AOB</math>. Through a point <math>M \neq O</math> perpendiculars are drawn to <math>OA</math> and <math>OB</math>, the feet of which are <math>P</math> and <math>Q</math> respectively. The point of intersection of the altitudes of <math>\triangle OPQ</math> is <math>H</math>. What is the locus of <math>H</math> if <math>M</math> is permitted to range over (a) the side <math>AB</math>, (b) the interior of <math>\triangle OAB</math>? | ||
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== Solution == | == Solution == | ||
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a line segment <math>MN , M \in OA , N \in OB</math>. | a line segment <math>MN , M \in OA , N \in OB</math>. | ||
Second question: the locus consists in the <math>\triangle OMN</math>. | Second question: the locus consists in the <math>\triangle OMN</math>. | ||
+ | |||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | This solution is a simplified version of the previous solution, | ||
+ | it fills in some gaps. and it provides more information. The | ||
+ | idea is to just follow the degrees of the expressions and | ||
+ | equations in <math>\lambda, x, y</math> involved. If we manage to conclude | ||
+ | that the equation for <math>H</math> is an equation of degree <math>1</math>, 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 <math>M</math> are expressions of degree <math>1</math> in <math>\lambda</math>. | ||
+ | |||
+ | The equation for <math>MP</math> is an equation of degree <math>1</math> in <math>x, y</math> | ||
+ | with constant coefficients for <math>x, y</math>, and whose constant term | ||
+ | is an expression of degree <math>1</math> in <math>\lambda</math>. | ||
+ | |||
+ | The coordinates of <math>P</math> (the intersection of <math>MP</math> and <math>OA</math>) are | ||
+ | expressions of degree <math>1</math> in <math>\lambda</math>. | ||
+ | |||
+ | The equation of the perpendicular from <math>P</math> to <math>OB</math> is of degree | ||
+ | <math>1</math> in <math>x, y</math>, with constant coefficients for <math>x, y</math>, and whose | ||
+ | constant term is an expression of degree <math>1</math> in <math>\lambda</math>. This | ||
+ | corresponds to equation (2) in the above solution. | ||
+ | |||
+ | Similarly, the equation of the perpendicular from <math>Q</math> to <math>OA</math> | ||
+ | is of degree <math>1</math> in <math>x, y</math>, with constant coefficients for | ||
+ | <math>x, y</math>, and whose constant term is an expression of degree | ||
+ | <math>1</math> in <math>\lambda</math>. This corresponds to equation (1) in the | ||
+ | above solution. | ||
+ | |||
+ | Now, in principle, we would have to solve the system of two | ||
+ | equations to obtain the coordinates of <math>H</math> as expressions | ||
+ | of <math>\lambda</math>, and then eliminate <math>\lambda</math> to obtain an | ||
+ | equation in <math>x, y</math>. Or, as a shortcut, we can eliminate | ||
+ | <math>\lambda</math> directly from the two equations. Either way, | ||
+ | the result is an equation of degree <math>1</math> in <math>x, y</math>. | ||
+ | |||
+ | This tells us that the locus is on this line. We just need to | ||
+ | specify which set of points on this line is the locus. | ||
+ | |||
+ | The previous solution, with a good amount of hand waving, tells | ||
+ | us that the solution is "a line segment | ||
+ | <math>B_1A_1, B_1 \in OA, A_1 \in OB</math>". (On top of the hand waving | ||
+ | the solution uses the unhappy notation <math>M</math> for <math>B_1</math> and <math>N</math> | ||
+ | for <math>A_1</math>, which is bad because <math>M</math> has already been used!) | ||
+ | We will do better than that. | ||
+ | |||
+ | 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>. | ||
+ | (For this paragraph see the picture shown in Solution 3.) | ||
+ | Consider the limit situation when <math>M = A</math>. Then <math>Q = A_1</math>, and | ||
+ | <math>P = A</math>. It follows that the intersection <math>H</math> of the | ||
+ | perpendiculars from <math>P</math> to <math>OB</math> and <math>Q</math> to <math>OA</math> is <math>A_1</math>. | ||
+ | Similarly, the limit situation when <math>M = B</math> yields <math>H = B_1</math>. | ||
+ | Now it is reasonable to say that when <math>M</math> moves from <math>A</math> to <math>B</math>, | ||
+ | <math>H</math> moves from <math>A_1</math> to <math>B_1</math>. So, the locus is the line segment | ||
+ | joining the feet <math>A_1, B_1</math> of the perpendiculars in | ||
+ | <math>\triangle OAB</math> from <math>A, B</math>. This answers question (a). | ||
+ | |||
+ | Again with a good amount of hand waving, the previous solution | ||
+ | says "the locus consists in the <math>\triangle OB_1A_1</math>". We justify | ||
+ | this by pointing out that if <math>M</math> is inside <math>\triangle OAB</math>, then | ||
+ | we can take the triangle <math>\triangle OA'B'</math>, such that <math>A' \in OA</math>, | ||
+ | <math>B' \in OB</math>, <math>A'B'</math> going through <math>M</math> and parallel to <math>AB</math>. Then | ||
+ | <math>H</math> will be on the corresponding segment <math>A_1'B_1'</math> determined by | ||
+ | the feet of the perpendiculars in <math>\triangle OA'B'</math>. Conversely, | ||
+ | it is easy to see that any point <math>H \in \triangle OA_1B_1</math> is on | ||
+ | a segment <math>A_1'B_1'</math> obtained from a triangle <math>\triangle OA'B'</math>, | ||
+ | and <math>H</math> is obtained from a point <math>M \in A'B'</math>. This answers | ||
+ | question (b). | ||
+ | |||
+ | (Solution by pf02, October 2024) | ||
+ | |||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | TO BE CONTINUED. SAVING MID WAY, SO I DON'T LOSE WORK DONE SO FAR. | ||
+ | |||
== 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 20:01, 30 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. 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 with constant coefficients for , and whose constant term is an expression of degree in .
The coordinates of (the intersection of and ) 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 to obtain the coordinates of as expressions of , and then eliminate to obtain an equation in . Or, as a shortcut, we can eliminate directly from the two equations. Either way, the result is an equation of degree in .
This tells us that the locus is on this line. We just need to specify which set of points on this line is the locus.
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).
Again 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
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 |