Difference between revisions of "1965 IMO Problems/Problem 5"
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== Solution == | == Solution == | ||
{{solution}} | {{solution}} | ||
+ | Let <math>O(0,0),A(a,0),B(b,c)</math>. | ||
+ | Equation of the line <math>AB: y=\frac{c}{b-a}(x-a)</math>. | ||
+ | Point <math>M \in AB : M(\lambda,\frac{c}{b-a}(\lambda-a))</math>. | ||
+ | Easy, point <math>P(\lambda,0)</math>. | ||
+ | Point <math>Q = OB \cap MQ</math>, <math>MQ \bot OB</math>. | ||
+ | Equation of <math>OB : y=\frac{c}{b}x</math>, equation of <math>MQ : y=-\frac{b}{c}(x-\lambda)+\frac{c}{b-a}(\lambda-a)</math>. | ||
+ | Solving: <math>x_{Q}=\frac{1}{b^{2}+c^{2}}\left[b^{2}\lambda+\frac{c^{2}(\lambda-a)b}{b-a}\right]</math>. | ||
+ | Equation of the first altitude: <math>x=\frac{1}{b^{2}+c^{2}}\left[b^{2}\lambda+\frac{c^{2}(\lambda-a)b}{b-a}\right] \quad (1)</math>. | ||
+ | Equation of the second altitude: <math>y=-\frac{b}{c}(x-\lambda)\quad\quad (2)</math>. | ||
+ | Eliminating <math>\lambda</math> from (1) and (2): | ||
+ | <cmath>ac \cdot x + (b^{2}+c^{2}-ab)y=abc </cmath> | ||
+ | a line segment <math>MN , M \in OA , N \in OB</math>. | ||
+ | Second question: the locus consists in the <math>\triangle OMN</math>. |
Revision as of 11:39, 1 June 2013
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
This problem needs a solution. If you have a solution for it, please help us out by adding it.
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
.