Difference between revisions of "2020 CIME I Problems/Problem 5"
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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>. | 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>. | ||
+ | ==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 12:25, 31 August 2020
Problem 5
Let be a rectangle with sides and let be the reflection of over . If and the area of is , find the area of .
Solution
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Let be the center of rectangle . Because is the reflection of over and degrees, we have degrees. This means lies on the circle with diameter , or the circumcircle of rectangle . We are given , so by symmetry, . Since the three lengths are equal and degrees, we must have degrees, so , , are all equilateral. Given that the area of cyclic quadrilateral is , the area of is . This is of the area of rectangle , so the answer is .
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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