# 2021 JMPSC Invitationals Problems/Problem 12

## 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

## Solution 2 (Trig)

Let $\angle DBC = \theta$. We have $\cos(90- \theta)=\sin(\theta)=\frac{7 \sqrt{65}}{65}$, and $\cos(\angle CDE)=\frac{4 \sqrt{65}}{65}$. Now, Law Of cosines on $\triangle DCE$ and $\triangle BCF$ gets $CF^2=25$ and $CE^2=58$, so $CE^2+CF^2=\boxed{83}.$ ~ Geometry285

## Solution 3 (Mass points and Ptolemy)

Let $O$ be the center of square $BDEF$. Applying moment of inertia to the system of mass points $\Sigma = {1B,1D,1E,1F}$ (which has center of mass $O$) gives $$CB^2 + CD^2 + CE^2 + CF^2 = OB^2 + OD^2 + OE^2 + OF^2 + 4OC^2.$$ Since $\triangle CBD$ is a right triangle, we may further cancel out some terms via Pythag to get $$CE^2 + CF^2 = OE^2 + OF^2 + 4OC^2 = 65 + 4OC^2.$$ To compute $OC$, apply Ptolemy to cyclic quadrilateral $DOCB$ (using the fact that $\triangle BOD$ is 45-45-90) to get $OC = \tfrac{3}{\sqrt 2}$. Thus $$CE^2 + CF^2 = 65 + 4\cdot\frac 92 = 65 + 18 = \boxed{83}.$$ ~djmathman

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