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

Latest revision as of 23:59, 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 $ADF$