1967 IMO Problems/Problem 4

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 $BC$ .

(Solution by pf02, September 2024)

TO BE CONTINUED. I AM SAVING MID WAY SO AS NOT TO LOSE WORK DONE SO FAR.

1967 IMO (Problems) • Resources
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 5
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1967 IMO Problems/Problem 4

Solution

Solution 2

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