Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 12"

Revision as of 17:35, 11 July 2021

Problem

Rectangle $ABCD$ is drawn such that $AB=7$ and $BC=4$ . $BDEF$ is a square that contains vertex $C$ in its interior. Find $CE^2+CF^2$ .

Solution 1 (Clever Construction)

We draw a line from $E$ to point $G$ on $DC$ such that $EG \perp CD$ . We then draw a line from $F$ to point $H$ on $EG$ such that $FH \perp EG$ . Finally, we extend $BC$ to point $I$ on $FH$ such that $CI \perp FH$ .

Next, if we mark $\angle CBD$ as $x$ , we know that $\angle BDC = 90-x$ , and $\angle EDG = x$ . We repeat this, finding $\angle CBD = \angle EDG = \angle FEH = \angle BFI = x$ , so by AAS congruence, $\triangle BDC \cong \triangle DEG \cong \triangle EFH \cong \triangle FBI$ . This means $BC = DG = EH = FI = AD = 4$ , and $DC = EG = FH = BI = AB = 7$ , so $CG = GH = HI = IC = 7-4 = 3$ . We see $CF^2 = CI^2 + IF^2 = 3^2 + 4^2 = 9 + 16 = 25$ , while $CE^2 = CG^2 + GE^2 = 3^2 + 7^2 = 9 + 49 = 58$ . Thus, $CF^2 + CE^2 = 25 + 58 = \boxed{83}.$ ~Bradygho

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 ==Solution 1 (Clever Construction)==
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Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 12"

Revision as of 17:35, 11 July 2021

Problem

Solution 1 (Clever Construction)

See also