We can draw a diagonal

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We can draw a diagonal

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#1 h Sep 30, 2010, 1:21 PM • 2 Y

Y by Adventure10, Mango247

Prove that in every convex hexagon of area $S$ one can draw a diagonal that cuts off a triangle of area not exceeding $\frac{1}{6}S.$

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TheChainheartMachine

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#2 h Apr 25, 2013, 1:28 AM • 1 Y

Y by Adventure10

Edit: What was I thinking? Was I even already fully awake when I posted this? Kyaaaah! *facepalm*

This post has been edited 1 time. Last edited by TheChainheartMachine, Apr 25, 2013, 6:45 AM

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#3 h Apr 25, 2013, 6:17 AM • 3 Y

Y by TheChainheartMachine, Adventure10, Mango247

^ What? Drawing all the diagonals from a vertex cuts the hexagon to four triangles only.

(FYI: TheChainheartMachine originally posted a solution along the lines of "diagonals from a vertex cut a hexagon to six triangles", that is now edited out.)

This post has been edited 1 time. Last edited by chaotic_iak, Apr 25, 2013, 11:42 AM

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Domi

#4 h Apr 25, 2013, 10:05 AM • 2 Y

Y by Adventure10, Mango247

First suppose that the main diagonals pass through one point and use pigeon hole .

Domi

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#5 h Apr 26, 2013, 6:50 AM • 2 Y

Y by Adventure10, Mango247

Amir Hossein wrote:

Prove that in every convex hexagon of area $S$ one can draw a diagonal that cuts off a triangle of area not exceeding $\frac{1}{6}S.$

It's obvious if the primary diagonals pass through one point, say O. Let the hexagon be $ABCDEI$ .
By primary diagonals, I mean diagonals that do not directly cut off a triangle, like AD,BE and CI.
Let $\triangle EOD$ be the triangle with the required property. Now one of the triangles $\triangle ECD$ or $\triangle EID$ has area less that $\triangle EOD$ and we are done.

Now, let the primary diagonals make a triangle FGH, with F&G on IC, H on BE & AD.
Look at the triangles - $\triangle ABF,BFC,GDC,AHI,EGD,EHI$
From the preceding principle there exists one triangle satisfying the given requirement.

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