Difference between revisions of "1965 IMO Problems/Problem 5"
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This solution is a simplified version of the previous solution, | This solution is a simplified version of the previous solution, | ||
− | and it provides more information | + | it fills in some gaps. and it provides more information. |
− | |||
− | |||
− | |||
− | |||
Just like in the previous solution, we use analytic (coordinate) | Just like in the previous solution, we use analytic (coordinate) | ||
geometry, but we don't care how the axes are chosen. | geometry, but we don't care how the axes are chosen. | ||
+ | |||
+ | Let <math>A, B, O</math> have coordinates <math>(a_1, a_2), (b_1, b_2), (c_1, c_2)</math>, | ||
+ | and let | ||
+ | <math>M = (\lambda a_1 + (1 - \lambda b_1, \lambda a_2 + (1 - \lambda b_2)</math> | ||
+ | with <math>\lambda \in [0, 1]</math>. | ||
+ | The idea is to just follow the degrees of the expressions and | ||
+ | equations in <math>\lambda, x, y</math> involved as we make the computations | ||
+ | for obtaining the coordinates of <math>H</math>, and the equation of the | ||
+ | curve <math>H</math> is on. We will see that the equation for <math>H</math> is an | ||
+ | equation of degree <math>1</math>, so we will know that it is a line. We | ||
+ | don't need to write out the equation explicitly. | ||
The coordinates of <math>M</math> are expressions of degree <math>1</math> in <math>\lambda</math>. | 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> | + | The equation for <math>MP</math> (the perpendicular from <math>M</math> to <math>OA</math>) is an |
− | with constant coefficients for <math>x, y</math>, and whose constant term | + | equation of degree <math>1</math> in <math>x, y</math> with constant coefficients for |
− | is an expression of degree <math>1</math> in <math>\lambda</math>. | + | <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 | + | The coordinates of <math>P</math> (the foot of the perpendicular from <math>P</math> to |
− | expressions of degree <math>1</math> in <math>\lambda</math>. | + | <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 | 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 | <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 | 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> | 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 | 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>x, y</math>, and whose constant term is an expression of degree | ||
− | <math>1</math> in <math>\lambda</math>. This | + | <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 (1) and (2) to obtain the coordinates of <math>H</math> as | ||
+ | expressions of <math>\lambda</math>, and then eliminate <math>\lambda</math> to | ||
+ | obtain the equation in <math>x, y</math> for <math>H</math>. As a shortcut, we | ||
+ | can eliminate <math>\lambda</math> directly from the two equations (1) | ||
+ | and (2). 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 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 | ||
+ | <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). | ||
+ | |||
+ | For part (b) of the problem, 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 == | ||
Latest revision as of 11:55, 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
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 |