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)

Invites12.png

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 \[CE^2 + CF^2 = OE^2 + OF^2 + 4OP^2 = 65 + 4OP^2.\] To compute $OP$, apply Ptolemy to cyclic quadrilateral $DOCB$ (using the fact that $\triangle BOD$ is 45-45-90) to get $OP = \tfrac{3}{\sqrt 2}$. Thus \[CE^2 + CF^2 = 65 + 4\cdot\frac 92 = 65 + 18 = \boxed{83}.\] ~djmathman

See also

  1. Other 2021 JMPSC Invitationals Problems
  2. 2021 JMPSC Invitationals Answer Key
  3. All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

Invalid username
Login to AoPS