Difference between revisions of "1974 IMO Problems/Problem 2"
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− | In the triangle ABC | + | ==Problem== |
− | is the geometric mean of AD and DB if and only if | + | In the triangle <math>ABC</math>, prove that there is a point <math>D</math> on side <math>AB</math> such that <math>CD</math> is the geometric mean of <math>AD</math> and <math>DB</math> if and only if <math>\sin{A}\sin{B} \leq \sin^2 (\frac{C}{2})</math>. |
− | <math>\sin{A}\sin{B} \leq \sin^2 (\frac{C}{2})</math>. | ||
==Solution== | ==Solution== | ||
+ | Let a point <math>D</math> on the side <math>AB</math>. | ||
+ | Let <math>CF</math> the altitude of the triangle <math>\triangle ABC</math>, and <math>C'</math> the symmetric point of <math>C</math> through <math>F</math>. | ||
+ | We bring a parallel line <math>L</math> from <math>C'</math> to <math>AB</math>. This line intersects the ray <math>CD</math> at the point <math>E</math>, and we know that <math>DE=DC</math>. | ||
− | + | The distance <math>d(L,AB)</math> between the parallel lines <math>L</math> and <math>AB</math> is <math>CF</math>. | |
− | + | Let <math>w = (O,R)</math> the circumscribed circle of <math>\triangle ABC</math>, and <math>MM'</math> the perpendicular diameter to <math>AB</math>, such that <math>M,C</math> are on difererent sides of the line <math>AB</math>. | |
− | + | In fact, the problem asks when the line <math>L</math> intersects the circumcircle. Indeed: | |
− | + | Suppose that <math>DC</math> is the geometric mean of <math>DA,DB</math>. | |
− | + | <math>DA \cdot DB = DC^{2}\Rightarrow DA \cdot DB = DC \cdot DE</math> | |
+ | |||
+ | Then, from the power of <math>D</math> we can see that <math>E</math> is also a point of the circle <math>w</math>. | ||
+ | Or else, the line <math>L</math> intersects <math>w \Leftrightarrow</math> | ||
+ | |||
+ | <math>d(L,AB)\leq d(M,AB) \Leftrightarrow</math> | ||
+ | |||
+ | <math>CF \leq MN,</math> where <math>MN</math> is the altitude of the isosceles <math>\triangle MAB</math>. | ||
+ | |||
+ | <math>\Leftrightarrow \frac{1}{2}CF \cdot AB \leq \frac{1}{2}MN \cdot AB \Leftrightarrow</math> <math>(ABC) \leq (MAB) \Leftrightarrow</math> <math>\frac{AB \cdot BC \cdot AC}{4R}\leq \frac{AB \cdot MA^{2}}{4R}\Leftrightarrow</math> | ||
+ | |||
+ | <math>BC \cdot AC \leq MA^{2}</math> | ||
+ | |||
+ | We use the formulas: | ||
+ | |||
+ | <math>BC = 2R \cdot \sin A</math> | ||
+ | <math>AC = 2R \cdot \sin B</math> | ||
+ | |||
+ | and <math>\angle CMA = \frac{C}{2}\Rightarrow MA = 2R \cdot \sin\frac{C}{2}</math> | ||
+ | |||
+ | So we have | ||
+ | <math>(2R \cdot \sin A)(2R \cdot \sin B) \leq (2R \cdot \sin\frac{C}{2})^{2}\Leftrightarrow</math> | ||
+ | <math>\sin A \cdot \sin B \leq \sin^{2}\frac{C}{2}</math> | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | For <math>(\Leftarrow)</math> | ||
+ | |||
+ | Suppose that <math>\sin A \cdot \sin B \leq \sin^{2}\frac{C}{2}</math> | ||
+ | |||
+ | Then we can go inversely and we find that <math>d(L,AB)\leq d(M,AB) \Leftrightarrow</math> | ||
+ | the line <math>L</math> intersects the circle <math>w</math> (without loss of generality; if <math>d(L,AB)=d(M,AB)</math> then <math>L</math> is tangent to <math>w</math> at <math>M</math>) | ||
+ | |||
+ | So, if <math>E \in L \cap w</math> then for the point <math>D = CE \cap AB</math> we have <math>DC=DE</math> and <math>AD \cdot AB = CD \cdot DE \Rightarrow</math> | ||
+ | |||
+ | <math>AD \cdot AB = CD^{2}</math> | ||
+ | |||
+ | The above solution was posted and copyrighted by pontios. The original thread for this problem can be found here: [https://aops.com/community/p365059] | ||
+ | |||
+ | == See Also == {{IMO box|year=1974|num-b=1|num-a=3}} |
Latest revision as of 15:56, 29 January 2021
Problem
In the triangle , prove that there is a point
on side
such that
is the geometric mean of
and
if and only if
.
Solution
Let a point on the side
.
Let
the altitude of the triangle
, and
the symmetric point of
through
.
We bring a parallel line
from
to
. This line intersects the ray
at the point
, and we know that
.
The distance between the parallel lines
and
is
.
Let the circumscribed circle of
, and
the perpendicular diameter to
, such that
are on difererent sides of the line
.
In fact, the problem asks when the line intersects the circumcircle. Indeed:
Suppose that is the geometric mean of
.
Then, from the power of we can see that
is also a point of the circle
.
Or else, the line
intersects
where
is the altitude of the isosceles
.
We use the formulas:
and
So we have
For
Suppose that
Then we can go inversely and we find that
the line
intersects the circle
(without loss of generality; if
then
is tangent to
at
)
So, if then for the point
we have
and
The above solution was posted and copyrighted by pontios. The original thread for this problem can be found here: [1]
See Also
1974 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |