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Difference between revisions of "1967 IMO Problems/Problem 4"
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This solution was posted and copyrighted by grobber. The thread can be found here: [https://aops.com/community/p139266]
 
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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[[File:Prob_1967_4_fig1.png|800px]]
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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 16:04, 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 $BC$.

Prob 1967 4 fig1.png


(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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