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Source: 1992 IrMO Paper 2 Problem 4

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984 posts

#1 h Oct 2, 2017, 4:24 AM • 2 Y

Y by Adventure10, Mango247

A convex pentagon has the property that each of its diagonals cuts off a triangle of unit area. Find the area of the pentagon.

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4894 posts

#2 h Oct 2, 2017, 10:55 AM • 2 Y

Y by Adventure10, Mango247

$S(AGB)=S(DGC) \to AG*GB=DG*GC \to \frac{AG}{DG}=\frac{GC}{GB} \to \triangle AGD \sim \triangle CGB \to AD\parallel BC$ . Also $\triangle AGB \sim \triangle DGC$ so $ABCD$ is cyclic. Same way we can prove, that $BCDE$ is cyclic, so $ABCDE$ is cyclic.
With simple angle chasing we can prove, that all triangles, cuted by diagonals, are equals with angles $108,36,36$

Then using trigonometry, we can find that $S(ABCDE)=2+\frac{\sqrt{5}+1}{2}$

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#3 h Oct 2, 2017, 10:57 AM • 1 Y

Y by Adventure10

https://artofproblemsolving.com/community/q1h1472860p8558005

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#4 h Oct 10, 2017, 5:57 AM • 1 Y

Y by Adventure10

RagvaloD wrote:

$S(AGB)=S(DGC) \to AG*GB=DG*GC \to \frac{AG}{DG}=\frac{GC}{GB} \to \triangle AGD \sim \triangle CGB \to AD\parallel BC$ . Also $\triangle AGB \sim \triangle DGC$ so $ABCD$ is cyclic. Same way we can prove, that $BCDE$ is cyclic, so $ABCDE$ is cyclic.
With simple angle chasing we can prove, that all triangles, cuted by diagonals, are equals with angles $108,36,36$

Then using trigonometry, we can find that $S(ABCDE)=2+\frac{\sqrt{5}+1}{2}$

What is $G$ ?

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6322 posts

#5 h Oct 11, 2017, 12:41 PM • 1 Y

Y by Adventure10

quangminh1173 wrote:

RagvaloD wrote:

$S(AGB)=S(DGC) \to AG*GB=DG*GC \to \frac{AG}{DG}=\frac{GC}{GB} \to \triangle AGD \sim \triangle CGB \to AD\parallel BC$ . Also $\triangle AGB \sim \triangle DGC$ so $ABCD$ is cyclic. Same way we can prove, that $BCDE$ is cyclic, so $ABCDE$ is cyclic.
With simple angle chasing we can prove, that all triangles, cuted by diagonals, are equals with angles $108,36,36$

Then using trigonometry, we can find that $S(ABCDE)=2+\frac{\sqrt{5}+1}{2}$

What is $G$ ?

I don't know what it is, but you can tell $AD\parallel BC$ because the altitudes are the same.

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727 posts

#6 h Oct 11, 2017, 12:58 PM • 1 Y

Y by Adventure10

Yes I've known that before. In fact, that what I used in my solution.

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