2015 AMC 12B Problem 13 — Solution | LIVE by Po-Shen Loh

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2015 AMC 12B Problem 13

Below is the professionally curated solution for Problem 13 of the 2015 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 12B solutions, or check the answer key.

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Concepts:cyclic quadrilateral inscribed angle isosceles triangle

Difficulty rating: 1670

13.

Quadrilateral $ABCD$ is inscribed in a circle with $\angle BAC = 70^\circ,$ $\angle ADB = 40^\circ,$ $AD = 4,$ and $BC = 6.$ What is $AC?$

$3 + \sqrt5$

$6$

$\dfrac92\sqrt2$

$8 - \sqrt2$

$7$

Solution:

Angles $BAC$ and $BDC$ subtend arc $BC,$ so $\angle BDC = 70^\circ.$ Then $\angle ADC = \angle ADB + \angle BDC = 110^\circ.$

Since $ABCD$ is cyclic, $\angle ABC = 180^\circ - 110^\circ = 70^\circ = \angle BAC.$ Thus $\triangle ABC$ is isosceles with $AC = BC = 6.$

Thus, the correct answer is B.

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