Difference between revisions of "2024 SSMO Accuracy Round Problems/Problem 5"

(Created page with "==Problem== Let <math>ABCD</math> be a convex quadrilateral such that <math>\angle ABC = 120^\circ</math> and <math>\angle ADC = 60^\circ</math>. If <math>AB = BC = CD = 5</m...")
 
 
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==Solution==
 
==Solution==
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From the Law of Cosines on triangle <math>ABC</math>, we have <cmath>AC^2 = 5^2+5^2-2\left(-\frac{1}{2}\right)\cdot5\cdot5\implies AC = 5\sqrt{3}.</cmath> Let <math>AD = x.</math> From the Law of Cosines on triangle <math>ADC,</math> we have <cmath>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.</cmath> Finally, we use the sine area formula: <cmath>[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}.</cmath>
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~SMO_Team

Latest revision as of 14:31, 10 September 2025

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