convex equilateral pentagon

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Source: Bundeswettbewerb Mathematik 1992, round two, problem 3

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orl

#1 h Jun 19, 2004, 9:42 AM • 2 Y

Y by Adventure10, Mango247

Provided a convex equilateral pentagon. On every side of the pentagon We construct equilateral triangles which run through the interior of the pentagon. Prove that at least one of the triangles does not protrude the pentagon's boundary.

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Myth

#2 h Dec 28, 2004, 1:42 PM • 2 Y

Y by Adventure10, Mango247

Fact 1. Pentagon $ABCDE$ contains at most 1 angle $<60^\circ$ .
Suppose contrary. It is clear that two angles $<60^\circ$ can't be two neighbour vertices. Consider Figure 1: $B,E<60^\circ$ . Then $AC,AD<1$ . Construct two points $C'$ and $D'$ s.t. $AC,AD<AC'=AD'<1$ . Since $\angle BAC,\angle EAD>60^\circ$ we conclude $\angle CAD<60^\circ$ , so $C'D'<1$ . So $CD$ lies inside triangle $AC'D'$ , whose sides $<1$ , so $CD<1$ . Contradiction.

Fact 2. Pentagon $ABCDE$ contains at most 2 angles $>120^\circ$ .
Suppose contary, then we can find two neighbour angles $>120^\circ$ . Consider Figure 2: $A,E>120^\circ$ . It immediately follows that projection $BD$ to line $AE$ is greater than 2, since $B'D'>1+\cos 60^\circ+\cos 60^\circ$ . Contradiction, since $B'D'\leq BD\leq 2$ .

Suppose pentagon $ABCDE$ doesn't contain angles, which is less than $60^\circ$ . Then, due to Fact 2, we can find two neighbour angles $\leq 120^\circ$ . Consider Figure 2. Suppose $A$ and $E$ such angles. Than $BB',DD'>\sin 60^\circ$ . It follows that all points of lines $BD$ are remote from $AE$ by at least $\sin 60^\circ$ . It means, that regular triangle constructed on side $AE$ lies inside pentagon.

Suppose pentagon $ABCDE$ contains angle $<60^\circ$ . Consider Figure 3: $A<60^\circ$ . We will show that $C,D\leq 120^circ$ . Indeed, suppose contary, i.e. $C>120^\circ$ . Than from triangle $EDC$ we obtain $EC>\sqrt{3}$ , on the other hand, it is necessarily $\angle EBC<120^\circ$ , so from triangle $EBC$ we obtain $EC<\sqrt 3$ (we use $EB<1$ ).
From Fact 1 we conclude $C,D\geq 60^\circ$ . So we can state that regular triangle constructed on $CD$ lies inside pentagon.

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