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

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Rectangle <math>ABCD</math> is drawn such that <math>AB=7</math> and <math>BC=4</math>. <math>BDEF</math> is a square that contains vertex <math>C</math> in its interior. Find <math>CE^2+CF^2</math>.
 
Rectangle <math>ABCD</math> is drawn such that <math>AB=7</math> and <math>BC=4</math>. <math>BDEF</math> is a square that contains vertex <math>C</math> in its interior. Find <math>CE^2+CF^2</math>.
  
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==Solution==
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==Solution 1 (Clever Construction)==
 
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[[File:Invites12.png|200px]]
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[[File:Invites12.png|500px]]
 
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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>.  
 
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>
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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> ~Bradygho
  
 
==See also==
 
==See also==

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)

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

See also

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

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