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Source: PRMO 2019 Leg 2 P6

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6107 posts

#1 h Aug 25, 2019, 4:45 PM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle such that $AB=AC$ . Suppose the tangent to the circumcircle of ABC at B is perpendicular to AC. Find angle ABC measured in degrees

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1915 posts

#2 h Aug 25, 2019, 5:06 PM • 2 Y

Y by mathingbingo, Adventure10

Was really easy.

Let the intersection of the tangent at $B$ and $AC$ be $K$ . Hence, $\angle KBA=\angle ACB=\theta(say)$ , also $\angle KAB=2\theta$ . Hence $\theta=\angle ABC=30^\circ$ .

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#3 h Aug 25, 2019, 5:12 PM • 2 Y

Y by Adventure10, Mango247

amar_04 wrote:

Was really easy.

Let the intersection of the tangent at $B$ and $AC$ be $K$ . Hence, $\angle KBA=\angle ACB=\theta(say)$ , also $\angle KAB=2\theta$ . Hence $\theta=\angle ABC=30^\circ$ .

Nice. Afterall, it was an actual 2 marker. Not to mention P1

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#4 h Aug 26, 2019, 5:06 AM • 1 Y

Y by Adventure10

In other words, $OB || AC$ . Which again gives a 120-30-30 triangle.

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#6 h Aug 31, 2019, 6:30 PM • 2 Y

Y by Ha_ha_ha, Adventure10

Is the tangent perpendicular to line segment AC or line containing AC?

And how is angle KBA = angle ACB?

Thank you.

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#7 h Nov 16, 2020, 3:06 PM • 4 Y

Y by Ha_ha_ha, Mango247, Mango247, Mango247

30 is the answer

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#8 h Nov 16, 2020, 11:25 PM • 1 Y

Y by Ha_ha_ha

Let the tangent at $B$ meet the extension of $CA$ at $X$ . Notice that $\measuredangle XBA =\measuredangle BCA =\measuredangle ACB$ . Considering $\triangle BXC$ , it is evident that the answer is $30^\circ$ .

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#9 h Jun 26, 2023, 8:05 AM

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will somebody show me a diagram for these question I am not getting it

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