Difference between revisions of "1967 IMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
− | + | We construct a point <math>P</math> inside <math>A_0B_0C_0</math> s.t. <math>\angle X_0PY_0=\pi-\angle X_1Z_1Y_1</math>, where <math>X,Y,Z</math> are a permutation of <math>A,B,C</math>. Now construct the three circles <math>\mathcal C_A=(B_0PC_0),\mathcal C_B=(C_0PA_0),\mathcal C_C=(A_0PB_0)</math>. We obtain any of the triangles <math>ABC</math> circumscribed to <math>A_0B_0C_0</math> and similar to <math>A_1B_1C_1</math> by selecting <math>A</math> on <math>\mathcal C_A</math>, then taking <math>B= AB_0\cap \mathcal C_C</math>, and then <math>B=CA_0\cap\mathcal C_B</math> (a quick angle chase shows that <math>B,C_0,A</math> are also colinear). | |
+ | We now want to maximize <math>BC</math>. Clearly, <math>PBC</math> always has the same shape (i.e. all triangles <math>PBC</math> are similar), so we actually want to maximize <math>PB</math>. This happens when <math>PB</math> is the diameter of <math>\mathcal C_B</math>. Then <math>PA_0\perp BC</math>, so <math>PC</math> will also be the diameter of <math>\mathcal C_C</math>. In the same way we show that <math>PA</math> is the diameter of <math>\mathcal C_A</math>, so everything is maximized, as we wanted. | ||
+ | This solution was posted and copyrighted by grobber. The thread can be found here: [https://aops.com/community/p139266] | ||
+ | |||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Since all the triangles <math>\triangle ABC</math> circumscribed to <math>\triangle A_0B_0C_0</math> | ||
+ | are similar, the one with maximum area will be the one with maximum sides, or | ||
+ | equivalently, the one with maximum side <math>BC</math>. So we will try to maximize <math>BC</math>. | ||
+ | |||
+ | The plan is to find the value of <math>\alpha</math> which maximizes <math>BC</math>. | ||
+ | |||
+ | [[File:Prob_1967_4_fig1.png|600px]] | ||
+ | |||
+ | Note that for any <math>\alpha</math> we can construct the line through <math>A_0</math> which | ||
+ | forms the angle <math>\alpha</math> with <math>A_0C_0</math>. We can construct points <math>B, C</math> | ||
+ | on this line, and lines through these points which form angles | ||
+ | <math>\angle B, \angle C</math> with the line, and which pass through <math>C_0, B_0</math> | ||
+ | respectively. Since <math>\triangle A_0B_0C_0, \triangle A_1B_1C_1</math> are acute, | ||
+ | <math>A_0</math> is between <math>B, C</math> and these lines will meet at a point <math>A</math> such that | ||
+ | <math>B_0</math> is between <math>A, C</math> and <math>C_0</math> is between <math>A, B</math>. | ||
+ | |||
+ | (More about this later.) | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | (Solution by pf02, September 2024) | ||
+ | |||
+ | TO BE CONTINUED. I AM SAVING MID WAY SO AS NOT TO LOSE WORK DONE SO FAR. | ||
+ | |||
+ | == See Also == | ||
+ | {{IMO box|year=1967|num-b=3|num-a=5}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
[[Category:Geometric Construction Problems]] | [[Category:Geometric Construction Problems]] |
Revision as of 17:21, 3 September 2024
Let and
be any two acute-angled triangles. Consider all triangles
that are similar to
(so that vertices
,
,
correspond to vertices
,
,
, respectively) and circumscribed about triangle
(where
lies on
,
on
, and
on
). Of all such possible triangles, determine the one with maximum area, and construct it.
Solution
We construct a point inside
s.t.
, where
are a permutation of
. Now construct the three circles
. We obtain any of the triangles
circumscribed to
and similar to
by selecting
on
, then taking
, and then
(a quick angle chase shows that
are also colinear).
We now want to maximize . Clearly,
always has the same shape (i.e. all triangles
are similar), so we actually want to maximize
. This happens when
is the diameter of
. Then
, so
will also be the diameter of
. In the same way we show that
is the diameter of
, so everything is maximized, as we wanted.
This solution was posted and copyrighted by grobber. The thread can be found here: [1]
Solution 2
Since all the triangles circumscribed to
are similar, the one with maximum area will be the one with maximum sides, or
equivalently, the one with maximum side
. So we will try to maximize
.
The plan is to find the value of which maximizes
.
Note that for any we can construct the line through
which
forms the angle
with
. We can construct points
on this line, and lines through these points which form angles
with the line, and which pass through
respectively. Since
are acute,
is between
and these lines will meet at a point
such that
is between
and
is between
.
(More about this later.)
(Solution by pf02, September 2024)
TO BE CONTINUED. I AM SAVING MID WAY SO AS NOT TO LOSE WORK DONE SO FAR.
See Also
1967 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |