Difference between revisions of "2020 AMC 12B Problems/Problem 18"

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Revision as of 22:27, 7 February 2020 (view source) Ogbooger (talk \| contribs) (Created page with "==Solution== Plot a point <math>F'</math> such that <math>F'</math> and <math>I</math> are collinear and extend line <math>FB</math> to point <math>B'</math> such that <math>...")		Latest revision as of 11:00, 9 May 2021 (view source) MRENTHUSIASM (talk \| contribs) (Fixed redirects so it is consistent to the AMC 10.) (Tag: New redirect)
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−	~~==Solution==~~	+	#redirect [[2020 AMC 10B Problems/Problem 21]]
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−	~~Plot a point <math>F'<~~/math> such that <math>F'</math> and <math>I</math> are collinear and extend line <math>FB</math> to point <math>B'</math> such that <math>FIB'F'</math> forms a square. Extend line <math>AE</math> to meet line F'B' and point <math>E'</math> is the intersection of the two. The area of this square is equivalent to <math>FI^2</math>. We see that the area of square <math>ABCD</math> is <math>4</math>, meaning each side is of length 2. The area of the quadrilateral <math>EIFF'E'</math> is <math>2</math>. Length <math>AE=\sqrt{2}</math>, thus <math>EB=2-\sqrt{2}</math>. Triangle <math>EB'E'</math> is isosceles, and the area of this triangle is <math>\frac{1}{2}(4-2\sqrt{2})(2-\sqrt{2})=6-4sqrt(2)</math>. Adding these two areas, we get $2+6-4\sqrt{2}=\boxed{\textbf{B) }8-4\sqrt{2}}