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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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==Solution 2==
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Since all the triangles <math>\triangle ABC</math> circumscribed to <math>\triangle A_0B_0C_0</math>
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are similar, the one with maximum area will be the one with maximum sides, or
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equivalently, the one with maximum side <math>BC</math>.  So we will try to maximize <math>BC</math>.
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The plan is to find the value of <math>\alpha</math> which maximizes <math>BC</math>.
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[[File:Prob_1967_4_fig1.png|600px]]
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Note that for any <math>\alpha</math> we can construct the line through <math>A_0</math> which
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forms the angle <math>\alpha</math> with <math>A_0C_0</math>.  We can construct points <math>B, C</math>
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on this line, and lines through these points which form angles
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<math>\angle B, \angle C</math> with the line, and which pass through <math>C_0, B_0</math>
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respectively.  Since <math>\triangle A_0B_0C_0, \triangle A_1B_1C_1</math> are acute,
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<math>A_0</math> is between <math>B, C</math> and these lines will meet at a point <math>A</math> such that
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<math>B_0</math> is between <math>A, C</math> and <math>C_0</math> is between <math>A, B</math>.
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(More about this later.)
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(Solution by pf02, September 2024)
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TO BE CONTINUED.  I AM SAVING MID WAY SO AS NOT TO LOSE WORK DONE SO FAR.
  
 
== See Also ==  
 
== See Also ==  

Revision as of 16:21, 3 September 2024

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]


Solution 2

Since all the triangles $\triangle ABC$ circumscribed to $\triangle A_0B_0C_0$ are similar, the one with maximum area will be the one with maximum sides, or equivalently, the one with maximum side $BC$. So we will try to maximize $BC$.

The plan is to find the value of $\alpha$ which maximizes $BC$.

Prob 1967 4 fig1.png

Note that for any $\alpha$ we can construct the line through $A_0$ which forms the angle $\alpha$ with $A_0C_0$. We can construct points $B, C$ on this line, and lines through these points which form angles $\angle B, \angle C$ with the line, and which pass through $C_0, B_0$ respectively. Since $\triangle A_0B_0C_0, \triangle A_1B_1C_1$ are acute, $A_0$ is between $B, C$ and these lines will meet at a point $A$ such that $B_0$ is between $A, C$ and $C_0$ is between $A, B$.

(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
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