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Difference between revisions of "2001 JBMO Problems/Problem 3"

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{{JBMO box|year=2001|num-b=2|num-a=4|five=}}

Revision as of 23:37, 8 January 2023

Problem 3

Let $ABC$ be an equilateral triangle and $D,E$ on the sides $[AB]$ and $[AC]$ respectively. If $DF,EG$ (with $F \in AE, G \in AD$) are the interior angle bisectors of the angles of the triangle $ADE$, prove that the sum of the areas of the triangles $DEF$ and $DEG$ is at most equal with the area of the triangle $ABC$. When does the equality hold?


Solution

We have \[\frac{[DEF]}{[DEA]} = \frac{EF}{EA} = \frac{DE}{DE+AD}\] Similarly \[\frac{[DEG]}{[DEA]} = \frac{DG}{DA} = \frac{DE}{DE+AE}\] Thus $[DEF] + [DEG] = DE[DEA](1/(DE+AD) + 1/(DE+AE))$. We have $[DEA] = 1/2 AD\cdot AE\sin A$, and $[ABC] = 1/2 AB\cdot AC \sin A$, so \[\frac{[DEF] + [DEG]}{[ABC]} = \frac{DE\cdot AD\cdot AE}{AB^2}\cdot\frac{1}{(DE+AD) + 1/(DE+AE)}\] Obviously $DE \le AB$, so it is sufficient to prove that $AD\cdot AE(1/(DE+AD) + 1/(DE+AE)) \le DE (*)$. The cosine rule applied to $ADE$ gives $DE^2 = AD^2 + AE^2 - AD\cdot AE$. Hence also $DE^2 - AD\cdot AE = (AD - AE)^2\ge 0$. Thus we have $DE(AD - AE)^2 + DE^2(AD + AE) \ge AD\cdot AE(AD + AE)$.

So \[AD\cdot AE(AD + AE + 2DE) \le DE(AD^2 + AE^2 + DE(AD + AE)) = DE(DE + AD)(DE + AE)\], which is $(*)$.


Bonus Question

In the above problem, prove that $DF/EG = AD/AE$.

- Proposed by $Kris17$

Solution to Bonus Question

Let $P$ be the intersection of $DF$ and $EG$, so $AP$ is an angle bisector of triangle $ADE$. Extend line DE to meet BC at H. Let us define $\angle DHB = \theta$

We have $\angle PDE = (60^{\circ} - \theta)/2 = (30^{\circ} - \theta/2)$ and $\angle PED = (60^{\circ} + \theta)/2 = (30^{\circ} + \theta/2)$ Thus $\angle DPE = 180^{\circ} - (\angle PDE + \angle PED) = 120^{\circ}$.

It follows that $AGPF$ is a cyclic quadrilateral. So we have $\angle GFP = \angle GAP = 30^{\circ}$, and $\angle PGF = \angle FAP = 30^{\circ}$

So $\angle GFP = \angle PGF$, implying that triangle $PGF$ is an isoceles triangle. So we have $PG = PF$.

Also, since $\angle GPD = 180^{\circ} - \angle DPE = 60^{\circ} = \angle DAF$, and $\angle GDP = \angle ADF$, it follows that: trianlge $GDP$ ~ triangle $ADF$

Similarly it can be shown that: trianlge $FEP$ ~ triangle $AEG$.

From trianlge $GDP$ ~ triangle $ADF$ we get: $PG/AF = DG/DF$, or $PG = AF(DG/DF)$

From triangle $FEP$ ~ triangle $AEG$ we get: $PF/AG = FE/GE$, or $PF = AG(FE/GE)$

Since $PG = PF$, we get $AF(DG/DF) = AG(FE/GE)$

or $(AF/FE)(DG/AG) = (DF/EG)$

or $(AD/DE)(DE/AE) = (DF/EG)$

Thus $DF/EG = AD/AE$


$Kris17$

2001 JBMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4
All JBMO Problems and Solutions


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