$A_0B_0C_0$ and $A_1B_1C_1$ are acute-angled triangles. Describe, and prove, how to construct the triangle $ABC$ with the largest possible area which is circumscribed about $A_0B_0C_0$ (so $BC$ contains $B_0, CA$ contains $B_0$, and $AB$ contains $C_0$) and similar to $A_1B_1C_1.$
Problem
Source: IMO LongList 1967, Italy 1
Tags: geometry, Triangle, similar triangles, area of a triangle, IMO, IMO 1967
21.12.2004 20:48
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. The acuteness of the triangles ensured that $P$ is inside $A_0B_0C_0$. I'm sure I used it in there somewhere .
31.12.2004 19:48
also of this problem, I remember, I already posted a solution some time ago. missing too? PS i have to check the grobber's solution: at first insigth, it seems equal to mine. If this is the case, I don't resend it. ciao buon anno!!!
31.12.2004 20:04
About lost solutions in the Geometry subforum, see http://www.mathlinks.ro/Forum/viewtopic.php?t=20496 . Happy New Year! Darij