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Difference between revisions of "1967 IMO Problems/Problem 4"
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==Solution==
 
==Solution==
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The solution to this problem can be found here: [https://artofproblemsolving.com/community/c6h21127p137262]
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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).
  
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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.
  
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This solution was posted and copyrighted by grobber. The thread can be found here: [https://aops.com/community/p139266]
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== See Also ==
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{{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 12:15, 29 January 2021

Let $A_0B_0C_0$ and $A_1B_1C_1$ be any two acute-angled triangles. Consider all triangles $ABC$ that are similar to $\triangle A_1B_1C_1$ (so that vertices $A_1$, $B_1$, $C_1$ correspond to vertices $A$, $B$, $C$, respectively) and circumscribed about triangle $A_0B_0C_0$ (where $A_0$ lies on $BC$, $B_0$ on $CA$, and $AC_0$ on $AB$). Of all such possible triangles, determine the one with maximum area, and construct it.


Solution

We construct a point $P$ inside $A_0B_0C_0$ s.t. $\angle X_0PY_0=\pi-\angle X_1Z_1Y_1$, where $X,Y,Z$ are a permutation of $A,B,C$. Now construct the three circles $\mathcal C_A=(B_0PC_0),\mathcal C_B=(C_0PA_0),\mathcal C_C=(A_0PB_0)$. We obtain any of the triangles $ABC$ circumscribed to $A_0B_0C_0$ and similar to $A_1B_1C_1$ by selecting $A$ on $\mathcal C_A$, then taking $B= AB_0\cap \mathcal C_C$, and then $B=CA_0\cap\mathcal C_B$ (a quick angle chase shows that $B,C_0,A$ are also colinear).

We now want to maximize $BC$. Clearly, $PBC$ always has the same shape (i.e. all triangles $PBC$ are similar), so we actually want to maximize $PB$. This happens when $PB$ is the diameter of $\mathcal C_B$. Then $PA_0\perp BC$, so $PC$ will also be the diameter of $\mathcal C_C$. In the same way we show that $PA$ is the diameter of $\mathcal C_A$, so everything is maximized, as we wanted.

This solution was posted and copyrighted by grobber. The thread can be found here: [1]


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
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