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Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 12"
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==Solution==
 
==Solution==
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We draw a line from <math>E</math> to point <math>G</math> on <math>DC</math> such that <math>EG \perp CD</math>. We then draw a line from <math>F</math> to point <math>H</math> on <math>EG</math> such that <math>FH \perp EG</math>. Finally, we extend <math>BC</math> to point <math>I</math> on <math>FH</math> such that <math>CI \perp FH</math>.
  
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Next, if we mark <math>\angle CBD</math> as <math>x</math>, we know that <math>\angle BDC = 90-x</math>, and <math>\angle EDG = x</math>. We repeat this, finding  <math>\angle CBD = \angle EDG = \angle FEH = \angle BFI = x</math>, so by AAS congruence, <math>\triangle BDC \cong \triangle DEG \cong \triangle EFH \cong \triangle FBI</math>. This means <math>BC = DG = EH = FI = AD = 4</math>, and <math>DC = EG = FH = BI = AB = 7</math>, so <math>CG = GH = HI = IC = 7-4 = 3</math>. We see <math>CF^2 = CI^2 + IF^2 = 3^2 + 4^2 = 9 + 16 = 25</math>, while <math>CE^2 = CG^2 + GE^2 = 3^2 + 7^2 = 9 + 49 = 58</math>. Thus, <math>CF^2 + CE^2 = 25 + 58 = \boxed{83}</math>
  
 
==See also==
 
==See also==

Revision as of 17:34, 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

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, $CF^2 + CE^2 = 25 + 58 = \boxed{83}$

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

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