2024 SSMO Accuracy Round Problems/Problem 5

Problem

Let $ABCD$ be a convex quadrilateral such that $\angle ABC = 120^\circ$ and $\angle ADC = 60^\circ$ . If $AB = BC = CD = 5$ , the area of $ABCD$ can be expressed as $a+b\sqrt{c},$ where $a,b,$ and $c$ are positive integers and $c$ is squarefree. Find $a+b+c.$

Solution

From the Law of Cosines on triangle $ABC$ , we have $AC^2 = 5^2+5^2-2\left(-\frac{1}{2}\right)\cdot5\cdot5\implies AC = 5\sqrt{3}.$ Let $AD = x.$ From the Law of Cosines on triangle $ADC,$ we have $x^2+5^2-2 \left(\frac12\right)5\cdot x = \left(5\sqrt{3}\right)^2\implies x^2-5x-50 = 0\implies x = 10.$ Finally, we use the sine area formula: $[ABCD] = [ABC]+[ACD] = \frac{1}{2}\left(5\cdot5\cdot\left(\frac{\sqrt{3}}{2}\right)+10\cdot5\left(\frac{\sqrt{3}}{2}\right)\right) = \frac{75\sqrt{3}}{4}\implies 75+3+4 = \boxed{82}.$

~SMO_Team

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2024 SSMO Accuracy Round Problems/Problem 5

Problem

Solution