Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 12"
(→‎Solution)
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
−
asdf
+
<center>
 +
[[File:Invites12.png|200px]]
 +
</center>
  
 +
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>.
  
 +
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 18:34, 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Invitationals_Problems/Problem_12&oldid=158131"