external bisector

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Source: 2021 Taiwan APMO Preliminary First Round

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#1 h Nov 21, 2020, 11:31 AM • 1 Y

Y by Mango247

$\triangle ABC$ , $\angle A=23^{\circ},\angle B=46^{\circ}$ . Let $\Gamma$ be a circle with center $C$ , radius $AC$ . Let the external angle bisector of $\angle B$ intersects $\Gamma$ at $M,N$ . Find $\angle MAN$ .

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282 posts

#2 h Nov 21, 2020, 2:40 PM

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I conjecture that $MACB$ is a cyclic quadrilateral, but I can't finish the problem with this fact alone. Interesting problem!

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#3 h Nov 21, 2020, 2:41 PM

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hsiangshen wrote:

$\triangle ABC$ , $\angle A=23^{\circ},\angle B=46^{\circ}$ . Let $\Gamma$ be a circle with center $C$ , radius $AC$ . Let the external angle bisector of $\angle B$ intersects $\Gamma$ at $M,N$ . Find $\angle MAN$ .

Angle chase/

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#4 h Nov 21, 2020, 2:59 PM • 1 Y

Y by Mango247

Abhaysingh2003 wrote:

hsiangshen wrote:

$\triangle ABC$ , $\angle A=23^{\circ},\angle B=46^{\circ}$ . Let $\Gamma$ be a circle with center $C$ , radius $AC$ . Let the external angle bisector of $\angle B$ intersects $\Gamma$ at $M,N$ . Find $\angle MAN$ .

Angle chase/

Thanks for the extremely enlightening and insightful solution. Indeed a wonderful addition to this post

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1455 posts

#5 h Nov 21, 2020, 3:52 PM • 1 Y

Y by Mango247

Let $C_{\perp}$ be the projection of $C$ on the internal angle bisector of $\angle B$ . Let $C'$ be the projection of $C$ on $MN$ .

Since $CC_{\perp}BC'$ is a rectangle (internal bisector $\perp$ external bisector), we have $CC' = BC_{\perp} = BC\cos(23 ^{\circ})$ .

By Pythagorean, we have $2R\sin(\angle MAN) = MN = 2MC' = 2\sqrt{AC^2-BC^2\cos^2(23^{\circ})}$

Simplifying, we have $\cos^2(\angle MAN) = \frac{BC^2}{AC^2} \cdot \cos^2(23^{\circ})$

Notice by LOS, we have $\frac{BC}{AC} = \frac{\sin(23^{\circ})}{\sin(46^{\circ})} = \frac{1}{2\cos(23^{\circ})}$

Thus, we have $\cos^2(\angle MAN) = \frac{1}{4}$ , and since $A$ is on the same side of $MN$ as $C$ (by definition, of external angle bisectors when $\angle ABC < 180^{\circ}$ ), we have $\angle MAN = 60^{\circ}$

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#6 h Nov 21, 2020, 4:33 PM • 1 Y

Y by Mango247

Inconsistent wrote:

Let $C_{\perp}$ be the projection of $C$ on the internal angle bisector of $\angle B$ . Let $C'$ be the projection of $C$ on $MN$ .

Since $CC_{\perp}BC'$ is a rectangle (internal bisector $\perp$ external bisector), we have $CC' = BC_{\perp} = BC\cos(23 ^{\circ})$ .

By Pythagorean, we have $2R\sin(\angle MAN) = MN = 2MC' = 2\sqrt{AC^2-BC^2\cos^2(23^{\circ})}$

Simplifying, we have $\cos^2(\angle MAN) = \frac{BC^2}{AC^2} \cdot \cos^2(23^{\circ})$

Notice by LOS, we have $\frac{BC}{AC} = \frac{\sin(23^{\circ})}{\sin(46^{\circ})} = \frac{1}{2\cos(23^{\circ})}$

Thus, we have $\cos^2(\angle MAN) = \frac{1}{4}$ , and since $A$ is on the same side of $MN$ as $C$ (by definition, of external angle bisectors when $\angle ABC < 180^{\circ}$ ), we have $\angle MAN = 60^{\circ}$

Nice solution! What led you to consider the rectangle?

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#7 h Nov 21, 2020, 4:47 PM

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Nice sol!
Looking forward to seeing pure geo sol

Most of my friends uses trig. too but I think there's a nice pure geo solution...

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#8 h Nov 21, 2020, 4:53 PM • 1 Y

Y by potatothegeek

potatothegeek wrote:

Nice solution! What led you to consider the rectangle?

We know the angle $\angle MAN$ is intrinsically tied to the length of $MN$ , which is tied to the distance of that line from $C$ , which is tied through trig to these projections. The rectangle was just there to make the trig ever so easier.

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#9 h Nov 21, 2020, 4:55 PM

Y by

potatothegeek wrote:

Abhaysingh2003 wrote:

hsiangshen wrote:

$\triangle ABC$ , $\angle A=23^{\circ},\angle B=46^{\circ}$ . Let $\Gamma$ be a circle with center $C$ , radius $AC$ . Let the external angle bisector of $\angle B$ intersects $\Gamma$ at $M,N$ . Find $\angle MAN$ .

Angle chase/

Thanks for the extremely enlightening and insightful solution. Indeed a wonderful addition to this post

You are welcome my dear!

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#10 h Nov 23, 2020, 1:36 AM • 3 Y

Y by Mango247, Mango247, Mango247

Ok so here's a solution from my friend Hakurei_Reimu
Let $AB$ intersects $\Gamma$ at $P$
We can easily see PERPENDICULAR! (And can be easily proved by angle chasing)
The rest is easy

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