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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=\frac{20\sqrt{2}}{\sqrt{41}}</math>, so <math>BD=\frac{40\sqrt{2}}{\sqrt{41}}=FE</math> by symmetry. Then, <math>FD=\frac{9\sqrt{2}}{\sqrt{41}}=FE</math> by the Pythagorean Theorem, so the area of <math>\Delta{DEF}</math> is <math>\frac{360}{41}</math>. Since <math>DA=FA=4\sqrt{2}</math> by symmetry, <math>FA=4\sqrt{2}</math>, so the area of <math>\Delta{FAD}</math> is <math>20</math>. This means that the area of the entire quadrilateral equals <math>\frac{360}{41}+20=28\frac{32}{41}</math>, so the answer is <math>28+32+41=\boxed{101}</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>
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~alexanderruan
 

Revision as of 02:10, 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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