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

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Let <math>ABC</math> be a triangle such that <math>AB=4\sqrt{2}, BC=5\sqrt{2},</math> and <math>AC=\sqrt{82}.</math> Let <math>\omega</math> be the circumcircle of <math>\triangle ABC</math>. Let <math>D</math> be on the circle such that <math>\overline{BD} \perp \overline{AC}.</math> Let <math>E</math> be the point diametrically opposite of <math>B</math>. Let <math>F</math> be the point diametrically opposite <math>D</math>. Find the area of the quadrilateral <math>ADEF</math> in terms of a mixed number <math>a\frac{b}{c}</math>. Find <math>a+b+c</math>.
 
Let <math>ABC</math> be a triangle such that <math>AB=4\sqrt{2}, BC=5\sqrt{2},</math> and <math>AC=\sqrt{82}.</math> Let <math>\omega</math> be the circumcircle of <math>\triangle ABC</math>. Let <math>D</math> be on the circle such that <math>\overline{BD} \perp \overline{AC}.</math> Let <math>E</math> be the point diametrically opposite of <math>B</math>. Let <math>F</math> be the point diametrically opposite <math>D</math>. Find the area of the quadrilateral <math>ADEF</math> in terms of a mixed number <math>a\frac{b}{c}</math>. Find <math>a+b+c</math>.
  
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==Solution==
+
==Solution 1==
 +
 
 +
Note that <math>\triangle ABC</math> is a right triangle by the Pythagorean Theorem.
 +
 
 +
Since <math>[ADF] = [ABC]</math>, we focus on computing <math>[ABC]</math> and <math>[BDEF]</math>. Let <math>O</math> be the midpoint of <math>\overline{AC}</math> (the center of the circumcircle), and let <math>T</math> be the foot of the perpendicular from <math>B</math> to <math>\overline{AC}</math>. Then
 +
<cmath>[BDEF] = 8 \cdot [BTO].</cmath>
 +
 
 +
We compute:
 +
<cmath>[ABC] = \frac{1}{2} \cdot 5\sqrt{2} \cdot 4\sqrt{2} = 20.</cmath>
 +
 
 +
So,
 +
<cmath>[ADEF] = [ADF] + [DEF] = [ABC] + 4 \cdot [BTO].</cmath>
 +
 
 +
Since <math>\triangle ABC \sim \triangle ATB</math>, we have
 +
<cmath>AT = AB \cdot \frac{AB}{AC} = \frac{32}{\sqrt{82}}.</cmath>
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Then,
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<cmath>TO = \frac{\sqrt{82}}{2} - \frac{32}{\sqrt{82}}, \quad \text{and} \quad BT = \frac{40}{\sqrt{82}}.</cmath>
 +
 
 +
Thus,
 +
<cmath>[BTO] = \frac{1}{2} \cdot \left( \frac{\sqrt{82}}{2} - \frac{32}{\sqrt{82}} \right) \cdot \frac{40}{\sqrt{82}} = \frac{90}{41}.</cmath>
 +
 
 +
The total area is:
 +
<cmath>[ADEF] = 20 + 4 \cdot \frac{90}{41} = 28\frac{32}{41},</cmath>
 +
so the final answer is
 +
<cmath>\boxed{101}.</cmath>
 +
 
 +
~SMO_Team
 +
 
 +
 
 +
==Solution 2==
  
 
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>.
 
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>.
  
 
~alexanderruan
 
~alexanderruan

Latest revision as of 21:16, 9 September 2025

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 1

Note that $\triangle ABC$ is a right triangle by the Pythagorean Theorem.

Since $[ADF] = [ABC]$, we focus on computing $[ABC]$ and $[BDEF]$. Let $O$ be the midpoint of $\overline{AC}$ (the center of the circumcircle), and let $T$ be the foot of the perpendicular from $B$ to $\overline{AC}$. Then \[[BDEF] = 8 \cdot [BTO].\]

We compute: \[[ABC] = \frac{1}{2} \cdot 5\sqrt{2} \cdot 4\sqrt{2} = 20.\]

So, \[[ADEF] = [ADF] + [DEF] = [ABC] + 4 \cdot [BTO].\]

Since $\triangle ABC \sim \triangle ATB$, we have \[AT = AB \cdot \frac{AB}{AC} = \frac{32}{\sqrt{82}}.\] Then, \[TO = \frac{\sqrt{82}}{2} - \frac{32}{\sqrt{82}}, \quad \text{and} \quad BT = \frac{40}{\sqrt{82}}.\]

Thus, \[[BTO] = \frac{1}{2} \cdot \left( \frac{\sqrt{82}}{2} - \frac{32}{\sqrt{82}} \right) \cdot \frac{40}{\sqrt{82}} = \frac{90}{41}.\]

The total area is: \[[ADEF] = 20 + 4 \cdot \frac{90}{41} = 28\frac{32}{41},\] so the final answer is \[\boxed{101}.\]

~SMO_Team


Solution 2

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

~alexanderruan

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