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Difference between revisions of "2020 CIME I Problems/Problem 5"
 
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==Solution==
 
==Solution==
 
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Let <math>O</math> be the center of rectangle <math>ABCD</math>. Because <math>E</math> is the reflection of <math>A</math> over <math>\overline{BD}</math> and <math>\angle BAD = 90</math> degrees, we have <math>\angle BED = 90</math> degrees. This means <math>E</math> lies on the circle with diameter <math>\overline{BD}</math>, or the circumcircle of rectangle <math>ABCD</math>. We are given <math>EC=BC</math>, so by symmetry, <math>EC=ED</math>. Since the three lengths are equal and <math>\overarc{AB}=180</math> degrees, we must have <math>\overarc{BC}=\overarc{CE}=</math>\overarc{ED}=60<math> degrees, so </math>\triangle OBC<math>, </math>\triangle OCE<math>, </math>\triangle OED<math> are all equilateral. Given that the area of cyclic quadrilateral </math>ECBD<math> is </math>144<math>, the area of </math>\triangle OBC<math> is </math>\frac{1}{3} \cdot 144 = 48<math>. This is </math>\frac{1}{4}<math> of the area of rectangle </math>ABCD<math>, so the answer is </math>48 \cdot 4 = \boxed{192}$.
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Let <math>O</math> be the center of rectangle <math>ABCD</math>. Because <math>E</math> is the reflection of <math>A</math> over <math>\overline{BD}</math> and <math>\angle BAD = 90</math> degrees, we have <math>\angle BED = 90</math> degrees. This means <math>E</math> lies on the circle with diameter <math>\overline{BD}</math>, or the circumcircle of rectangle <math>ABCD</math>. We are given <math>EC=BC</math>, so by symmetry, <math>EC=ED</math>. Since the three lengths are equal and <math>\overarc{BD}=180</math> degrees, we must have <math>\overarc{BC}=\overarc{CE}=\overarc{ED}=60</math> degrees, so <math>\triangle OBC</math>, <math>\triangle OCE</math>, <math>\triangle OED</math> are all equilateral. Given that the area of cyclic quadrilateral <math>ECBD</math> is <math>144</math>, the area of <math>\triangle OBC</math> is <math>\frac{1}{3} \cdot 144 = 48</math>. This is <math>\frac{1}{4}</math> of the area of rectangle <math>ABCD</math>, so the answer is <math>48 \cdot 4 = \boxed{192}</math>.
  
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==See also==
 
{{CIME box|year=2020|n=I|num-b=4|num-a=6}}
 
{{CIME box|year=2020|n=I|num-b=4|num-a=6}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 
{{MAC Notice}}
 
{{MAC Notice}}

Latest revision as of 11:25, 31 August 2020

Problem 5

Let $ABCD$ be a rectangle with sides $AB>BC$ and let $E$ be the reflection of $A$ over $\overline{BD}$. If $EC=AD$ and the area of $ECBD$ is $144$, find the area of $ABCD$.

Solution

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

Let $O$ be the center of rectangle $ABCD$. Because $E$ is the reflection of $A$ over $\overline{BD}$ and $\angle BAD = 90$ degrees, we have $\angle BED = 90$ degrees. This means $E$ lies on the circle with diameter $\overline{BD}$, or the circumcircle of rectangle $ABCD$. We are given $EC=BC$, so by symmetry, $EC=ED$. Since the three lengths are equal and $\overarc{BD}=180$ degrees, we must have $\overarc{BC}=\overarc{CE}=\overarc{ED}=60$ degrees, so $\triangle OBC$, $\triangle OCE$, $\triangle OED$ are all equilateral. Given that the area of cyclic quadrilateral $ECBD$ is $144$, the area of $\triangle OBC$ is $\frac{1}{3} \cdot 144 = 48$. This is $\frac{1}{4}$ of the area of rectangle $ABCD$, so the answer is $48 \cdot 4 = \boxed{192}$.

See also

2020 CIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All CIME Problems and Solutions

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