equal circumcircles in regular dodecagon, GERMANYISHOT

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equal circumcircles in regular dodecagon, GERMANYISHOT

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Source: 2009 Swiss IMO TST p1

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#1 h May 22, 2020, 6:33 PM • 2 Y

Y by Mango247, Mango247

Let $GERMANYISHOT$ be a regular dodecagon and let $P$ be the intersection point of $GN$ and $MI$ . Prove that:
(a) the circumscribed circle of the triangle $GIP$ has the same size as the circumscribed circle of $GERMANYISHOT$ .
(b) the $PA$ segment has the same length as the side of $GERMANYISHOT$ .

This post has been edited 1 time. Last edited by parmenides51, Jun 21, 2022, 1:49 AM

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#2 h May 23, 2020, 6:24 AM • 1 Y

Y by Atpar

Let the radius of circumscribed circle of $GERMANYISHOT$ be $R$
Let the radius of circumscribed circle of $\triangle GIP$ be $r$

Therefore, $2r=\frac{GI}{sin \angle GPI} , 2R=\frac{GI}{sin \angle GMI}$

Angle subtended by at the circumscribed circle by a side of dodecagon is $15^\circ$
Hence, $\angle NIG=30^\circ , \angle GIM=45^\circ , \angle GMI=75^\circ \implies \angle GPI=105^\circ$
$sin 105^\circ = sin 75^\circ \implies sin \angle GPI=sin \angle GMI \implies r=R$
Hence, the circumscribed circle of the triangle $GIP$ has the same size as the circumscribed circle of $GERMANYISHOT$
$\angle PNA=\angle GNA=60^\circ$
$\angle PIN=\angle MIN=30^\circ , \angle IPN=75^\circ$

Hence, $\frac{sin 30^\circ}{PN}=\frac{sin 75^\circ}{IN}$
But $IN=2Rsin 30^\circ \implies PN=2R \frac{sin^2 30^\circ}{sin 75^\circ}$

$NA$ is side length $\implies NA=2R sin 15^\circ$

$sin 15 ^\circ . sin 75^\circ = sin^2 30^\circ \implies NA=PN$
As $\angle PNA=60^\circ$ and $NA=PN$ , $\triangle PNA$ is equilateral
Hence, $PA=NA \implies PA$ segment has the same length as the side of $GERMANYISHOT$

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