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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>, both of which are right triangles. Mark the intersection point between BD and AC as G. We can use the fact that <math>\Delta{ABC}</math> and <math>\Delta{AGB}</math> are similar to find that <math>BG=20\sqrt{2}/\sqrt{41}</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>, both of which are right triangles. Mark the intersection point between BD and AC as G. We can use the fact that <math>\Delta{ABC}</math> and <math>\Delta{AGB}</math> are similar to find that <math>BG=\frac{20\sqrt{2}}{\sqrt{41}}</math>

Revision as of 00:59, 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 $\Delta{FAD}$, both of which are right triangles. Mark the intersection point between BD and AC as G. We can use the fact that $\Delta{ABC}$ and $\Delta{AGB}$ are similar to find that $BG=\frac{20\sqrt{2}}{\sqrt{41}}$

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