IMO LongList 1989, problem 2, Australia 2, shortlisted

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IMO LongList 1989, problem 2, Australia 2, shortlisted

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Source: IMO 1989/2 , ISL 1, ILL 2

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orl

#1 h Apr 4, 2005, 7:46 PM • 3 Y

Y by mathematicsy, Adventure10, Mango247

$ABC$ is a triangle, the bisector of angle $A$ meets the circumcircle of triangle $ABC$ in $A_1$ , points $B_1$ and $C_1$ are defined similarly. Let $AA_1$ meet the lines that bisect the two external angles at $B$ and $C$ in $A_0$ . Define $B_0$ and $C_0$ similarly. Prove that the area of triangle $A_0B_0C_0 = 2 \cdot$ area of hexagon $AC_1BA_1CB_1 \geq 4 \cdot$ area of triangle $ABC$ .

This post has been edited 1 time. Last edited by orl, Sep 18, 2008, 12:55 PM

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#2 h Apr 4, 2005, 9:42 PM • 2 Y

Y by Adventure10, Mango247

I'll post a solution to the second part (the inequality) for now:

Observe that 2[ACB1] = R * b * (1 - cosB), where R is the circumradius. Hence it suffices to show that p >= a * cosA + b * cosB + c * cosC, where p is the semi-perimeter. Now it's not difficult to convert this into proving: 2p / R >= sin2A + sin2B + sin2C = 4sinA * sinB * sinC. Use sine law to reduce it to: 2p * R<sup>2</sup> >= abc, then the result follows immediately from 4R = abc / S, S = rp, and R >= 2r, where S and r are the area and inradius of ABC, respectively.

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probability1.01

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probability1.01

#3 h Jul 26, 2006, 1:17 AM • 2 Y

Y by Adventure10, Mango247

Note that $ABC$ is the orthic triangle of $A_{0}B_{0}C_{0}$ , so it lies on the nine-point circle, and since $A_{1}B_{1}C_{1}$ is a homothety of $A_{0}B_{0}C_{0}$ about the incenter $I$ , we further have that the ratio must be one half (as the nine-point circle has half the radius of the circumcircle). It is then clear that $[IA_{1}BC_{1}] = \frac{1}{2}\cdot [A_{0}IC_{0}]$ . Adding these analogous equations yields the first equality.

For the second part, it suffices to show that $2[ABC] \le [BA_{1}CB_{1}AC_{1}]$ . Indeed, if we let $P, Q, R$ be points on the circumcircle with $AP \perp BC$ , etc., then $[BA_{1}CB_{1}AC_{1}] \ge [ARBPCQ] = 2 [ABC]$ .

If for example angle A is obtuse, then we already have $[ABA_{1}C] \ge 2[ABC]$ .

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#4 h Jul 2, 2014, 7:22 PM • 2 Y

Y by Adventure10, Mango247

This is actually length bashable. Use the $\frac{1}{2} ab\sin C$ formula to find the areas of triangles ABC1 and the linik, and then just bash bash bash untnil you get something like $\sum_{cyc} 2abc^2-a^2c^2-b^2c^2+c^4 \ge (a+b+c)(a+b-c)(a+c-b)(b+c-a)$ , which is easily provable by Schur's then AM-GM. (hahaha Muirhead fails...)

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