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Difference between revisions of "2024 SSMO Speed Round Problems/Problem 5"
 
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Note that quadrilateral <math>ABCD</math> is a cyclic quadrilateral with diameter <math>AC.</math> Since <math>\angle{ACB} = 30^\circ</math> and <math>BC = 3\sqrt{3},</math> we have <math>AC = \frac{BC}{\cos\angle{ACB}} = \frac{3\sqrt{3}}{\frac{\sqrt{3}}{2}} = 6.</math> So, the maximum possible length of <math>BD</math> is <math>6,</math> as the maximum length of any chord on a circle is the diameter.  
 
Note that quadrilateral <math>ABCD</math> is a cyclic quadrilateral with diameter <math>AC.</math> Since <math>\angle{ACB} = 30^\circ</math> and <math>BC = 3\sqrt{3},</math> we have <math>AC = \frac{BC}{\cos\angle{ACB}} = \frac{3\sqrt{3}}{\frac{\sqrt{3}}{2}} = 6.</math> So, the maximum possible length of <math>BD</math> is <math>6,</math> as the maximum length of any chord on a circle is the diameter.  
  
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SMO_Team
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~SMO_Team

Latest revision as of 14:27, 10 September 2025

Problem

Let $\triangle ABC$ and $\triangle ADC$ be right triangles, such that $\angle ABC = \angle ADC = 90^\circ$. Given that $\angle ACB = 30^\circ$ and $BC = 3\sqrt{3}$, find the maximum possible length of $BD$.

Solution

Note that quadrilateral $ABCD$ is a cyclic quadrilateral with diameter $AC.$ Since $\angle{ACB} = 30^\circ$ and $BC = 3\sqrt{3},$ we have $AC = \frac{BC}{\cos\angle{ACB}} = \frac{3\sqrt{3}}{\frac{\sqrt{3}}{2}} = 6.$ So, the maximum possible length of $BD$ is $6,$ as the maximum length of any chord on a circle is the diameter.

~SMO_Team

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