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Difference between revisions of "2023 SSMO Team Round Problems/Problem 3"

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==Solution==
 
==Solution==
  
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Note that <math>\Delta{ABC}</math> is right with the right angle at <math>B</math>. This means that <math>AC</math> is the diameter of the circle. We can divide quadrilateral <math>ADEF</math> into <math>\Delta{DEF}</math> and <math>\Delta{FAD}</math>
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Note that <math>\Delta{ABC}</math> is right with the right angle at <math>B</math>. This means that <math>AC</math> is the diameter of the circle. We can divide quadrilateral <math>ADEF</math> into <math>\Delta{DEF}</math> and <math>\Delta{FAD}</math>.

Revision as of 00:55, 3 January 2024

Problem

Let $ABC$ be a triangle such that $AB=4\sqrt{2}, BC=5\sqrt{2},$ and $AC=\sqrt{82}.$ Let $\omega$ be the circumcircle of $\triangle ABC$. Let $D$ be on the circle such that $\overline{BD} \perp \overline{AC}.$ Let $E$ be the point diametrically opposite of $B$. Let $F$ be the point diametrically opposite $D$. Find the area of the quadrilateral $ADEF$ in terms of a mixed number $a\frac{b}{c}$. Find $a+b+c$.

Solution

Note that $\Delta{ABC}$ is right with the right angle at $B$. This means that $AC$ is the diameter of the circle. We can divide quadrilateral $ADEF$ into $\Delta{DEF}$ and $\Delta{FAD}$.

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