For a convex hexagon $ ABCDEF$ with an area $ S$, prove that: \[ AC\cdot(BD+BF-DF)+CE\cdot(BD+DF-BF)+AE\cdot(BF+DF-BD)\geq 2\sqrt{3}S \]
Problem
Source: International Zhautykov Olympiad 2009, day 1, problem 3.
Tags: inequalities, geometry, trigonometry, trig identities, Law of Cosines, geometry proposed
je4ko
21.01.2009 14:11
very nice and difficult probem. in the official solution there is proved a lemma which I didn't know and I didn't solve the problem but it's nice and learnable
Erken
21.01.2009 14:12
@je4ko:and what did you get?
je4ko
21.01.2009 20:46
silver . I could have 25 point without any problem
thaithuan_GC
29.01.2009 10:12
I am very happy if someone gives a solution or hint. Thanks.
Erken
29.01.2009 15:11
thaithuan_GC wrote: I am very happy if someone gives a solution or hint. Thanks. no, let's better wait till someone else will come up with own solution. I did solve the problem, but it is approximately the same as the official, actually, the lemma used there was in one of our TST's.
silversheep
08.02.2009 21:43
Lemma. Let the incircle of BDF touch DF, FB, BD at X, Y, Z respectively. Then there exists a point O such that $ OB \leq \frac {2}{\sqrt {3}} BY$, $ OD \leq \frac {2}{\sqrt {3}} DZ$, $ OF \leq \frac {2}{\sqrt {3}} FX$. Proof.
Let u=DZ, v=FX, t=BY. It's not hard to see that there exists r such that the circles centered at B, D, F with radii rt, ru, rv have a point in common. Call this point O. We claim $ r \leq \frac {2}{\sqrt {3}}$. Since $ \angle BOD + \angle DOF + \angle FOB = 360$ one of these angles is at least 120, say without loss of generality $ \angle FOD$. Then using the Law of Cosines on $ \Delta FOD$,
\begin{eqnarray*} \frac { - 1}{2} = \cos 120 \geq \cos \angle FOD & = & \frac {OF^2 + OD^2 - FD^2}{2OF\cdot OD} \\
& = & \frac {r^2 v^2 + r^2 u^2 - (u + v)^2}{2uvr^2} \end{eqnarray*}
Now if $ r > \frac {2}{\sqrt {3}}$ this rearranges to $ 2uv > u^2 + v^2$ contradicting the Arithmetic Mean- Geometric Mean inequality. So $ r \leq \frac {2}{\sqrt {3}}$ and O satisfies the conditions.
Now,
$ \begin{eqnarray*} AC \cdot (BD + BF - DF) & = & 2 AC \cdot BZ \\
& \geq & 2 AC \cdot \frac {\sqrt {3}}{2} BO \\
& \geq & 2 \sqrt {3}\left[ABCO\right]$.
Adding up similar inequalities gives the result.
Some notes on motivation:
1. The BD+BF-DF, etc. in the problem reminds me of the length of a tangent to the incircle, so I rewrite it with BZ, etc.
2. I would very much like the left-hand side to be a sum of areas that add to the area of the hexagon. This gives motivation to somehow come up with a point O such that $ AC(BD + BF - DF) \geq 2\sqrt {3} \left[ABCO\right]$; then we could add cyclically.
3. The area of ABCO is at most $ BO \cdot AC$. So this motivates us to come up with the relation in the lemma.
jgnr
19.03.2009 15:10
silversheep wrote: 3. The area of ABCO is at most $ BO \cdot AC$. So this motivates us to come up with the relation in the lemma. is this also true for non-convex quadrilateral? how to prove it?
silversheep
19.03.2009 18:53
Quote: 3. The area of ABCO is at most $ BO \cdot AC$. So this motivates us to come up with the relation in the lemma. is this also true for non-convex quadrilateral? how to prove it? Suppose ABCO is oriented counterclockwise. We use directed angles and areas (counterclockwise is positive). Let BO and AC intersect at P. We have, using A=1/2bh, $ [ABC]=\frac{1}{2} AC \cdot BP \sin(\angle APB)$ and $ [COA]=\frac{1}{2} AC \cdot OP \sin(\angle CPO)$ Add these together and use that sin is less than 1.
jgnr
20.03.2009 13:54
thanks for your reply silversheep, i didn't realize that obvious thing. silversheep wrote: It's not hard to see that there exists r such that the circles centered at B, D, F with radii rt, ru, rv have a point in common. i cannot 'see' it, can you prove it?
silversheep
25.03.2009 18:46
Quote: It's not hard to see that there exists r such that the circles centered at B, D, F with radii rt, ru, rv have a point in common. i cannot 'see' it, can you prove it? Consider circles centered at B, D, F with radii rt, ru, and rv. As r increases from 0, the circles expand so eventually there will be a point in XYZ on the boundary or in the interior of all three circles. Take the least r such that this is true. Then if all three circles do not have a point in common, there is a point P strictly inside all three circles. Decreasing r by a sufficiently small amount P will still be inside all three circles, contradicting the minimality of r.
samirka259
08.01.2016 01:48
silversheep wrote: 3. The area of ABCO is at most $ BO \cdot AC$. So this motivates us to come up with the relation in the lemma. Actually, its area is at most $\frac{BO \cdot AC}{2}$ since $max(sin \alpha)=1$, isn't it ?
Marinchoo
18.12.2023 23:33
My solution is pretty much the same as the one in #7, so I'll just expand on the magical point $X$ (and how I came to construct it). Rather than intersecting circles centered at $B$, $D$, and $F$ with radii $rt$, $ru$, $rv$, consider the Apollonius circles of $DF$, $FB$, $BD$ going through $X$, $Y$, and $Z$, respectively. To prove these concur at one point, take the internal intersection of two of these circles, and using the equal ratios, it's clear that this point lies on the third Apollonius circle as well.