Prove that angle BAC is greater than 90°

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Prove that angle BAC is greater than 90°

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Source: CGMO 2003, Problem 7

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orl

#1 h Dec 28, 2008, 11:19 PM • 1 Y

Y by Adventure10

Let the sides of a scalene triangle $\triangle ABC$ be $AB = c,$ $BC = a,$ $CA =b,$ and $D, E , F$ be points on $BC, CA, AB$ such that $AD, BE, CF$ are angle bisectors of the triangle, respectively. Assume that $DE = DF.$ Prove that

(1) $\frac{a}{b+c} = \frac{b}{c+a} + \frac{c}{a+b}$

(2) $\angle BAC > 90^{\circ}.$

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yetti

#2 h Jan 1, 2009, 6:58 AM • 3 Y

Y by Arab, Adventure10, Mango247

Circumcircle $\mathcal P$ of $\triangle DEF$ cuts line $AD$ again at a point $A'.$ Since $DE = DF,$ line $AD$ bisects both the angles $\measuredangle EAF = \measuredangle CAB$ and $\measuredangle EA'F.$ This is possible only when $AD \perp EF,$ which means that $\triangle AFE, \triangle ABC$ are A-isosceles, or when points $A' \equiv A$ are identical, which means that quadrilateral $AFDE$ is cyclic. The first possibility is eliminated by the condition that $\triangle ABC$ is scalene.
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Circumcircle $\mathcal P$ of $AFDE$ cuts $BC$ at $D$ and again at a point $D'.$ Denote $d = \overline{DD'}.$ On one hand, powers of $B, C$ to $\mathcal P$ are

$\text{pow} (B \to \mathcal P) = \overline{BA} \cdot \overline {BF} = c \cdot \frac {ca}{a + b},$ $\quad \text{pow} (C \to \mathcal P) = \overline{CA} \cdot \overline {CE} = b \cdot \frac {ab}{c + a}.$

On the other hand,

$\text{pow} (B \to \mathcal P) = \overline{BD} \cdot \overline {BD'} = \frac {ca}{b + c} \cdot \left(\frac {ca}{b + c} \pm d\right),$ $\quad \text{pow} (C \to \mathcal P) = \overline{CD} \cdot \overline {CD'} = \frac {ab}{b + c} \cdot \left(\frac {ab}{b + c} \mp d\right),$

where upper signs apply for $b < c$ and lower signs for $b > c$ .
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Comparing the two expressions for $\text{pow} (B \to \mathcal P)$ and the two expressions for $\text{pow} (C \to \mathcal P)$ $\implies$

$\frac{ca}{b + c} \cdot \left(\frac {ca}{b + c} \pm d\right) = c \cdot \frac {ca}{a + b},$ $\quad \frac {ab}{b + c} \cdot \left(\frac {ab}{b + c} \mp d\right) = b \cdot \frac {ab}{c + a}.$

Reducing these equations by factors . $ca,\ ab,$ .resp. $\implies$

$\frac{1}{b + c} \cdot \left( \frac {ca}{b + c} \pm d \right) = \frac {c}{a + b},$ $\quad \frac{1}{b + c} \cdot \left( \frac {ab}{b + c} \mp d \right) = \frac {b}{c + a}.$

Adding the two equations to eliminate segment $d$ $\implies$

$\frac{1}{b + c} \cdot \left( (c + b) \cdot \frac {a}{b + c} \right) = \frac {c}{a + b} + \frac {b}{c + a}.$

Reducing left side of the last equation by factor $(b + c) = (c + b)$ $\implies$

$\boxed{ \frac {a}{b + c} = \frac {b}{c + a} + \frac {c}{a + b}} \ .$
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Multiplying this through $\implies$ (details)

$a^2c + a^3 + abc + a^2b = b^2a + b^3 + abc + b^2c + c^2b + c^3 + abc + c^2a.$

Collecting terms with factors . $a^2,\ b^2,\ c^2$ $\implies$

$a^2(c + a + b) + abc = b^2(a + b + c) + abc + c^2(b + c + a) + abc,$

$a^2 = b^2 + c^2 + \frac {abc}{a + b + c} > b^2 + c^2.$

By reverse Pythagorean theorem, angle $\boxed {\measuredangle CAB > 90^\circ}$ is obtuse.
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This post has been edited 27 times. Last edited by yetti, Nov 4, 2016, 6:03 AM
Reason: Adding details to the calculation. Improving LaTeX and language.

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#3 h Jul 3, 2013, 11:01 AM • 2 Y

Y by Adventure10, Mango247

it can be done using cosines lemma in $\triangle CED$ and $\triangle BFD$ on $DE,DF$ ,respectively.
and of course doing a lot of calculations.
(the solution isn't mine,but i have a sketch of it's proof in this way)

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#4 h Sep 30, 2017, 4:01 PM • 3 Y

Y by tenplusten, Adventure10, Mango247

See the proposed problem P5 from here.

This post has been edited 1 time. Last edited by Virgil Nicula, Sep 30, 2017, 4:01 PM

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