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 "1968 AHSME Problems/Problem 6"

m (→‎See also)
(→‎Solution)
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{E}</math>
+
Because <math>ABCD</math> is a convex [[quadrilateral]], the sum of its interior angles is <math>360^{\circ}</math>. Thus, <math>S'=\measuredangle BAD + \measuredangle ABC=360^{\circ}-(\measuredangle BCD+\measuredangle ADC)</math>. Furthermore, because <math>\angle BCD</math> and <math>\angle ADC</math> are [[supplementary]] to <math>\angle DCE</math> and <math>\angle CDE</math>, respectively, the four angles sum to <math>2*180^{\circ}=360^{\circ}</math>, so <math>\measuredangle BCD+\measuredangle ADC=360^{\circ}-(\measuredangle DCE+\measuredangle CDE)=360^{\circ}-S</math>. Plussing this expression for <math>\measuredangle BCD+\measuredangle ADC</math> into the first equation, we see that <math>S'=360^{\circ}-(360^{\circ}-S)=S</math>, so <math>r=\frac{S}{S'}=1</math>, which is answer choice <math>\fbox{E}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 17:34, 17 July 2024

Problem

Let side $AD$ of convex quadrilateral $ABCD$ be extended through $D$, and let side $BC$ be extended through $C$, to meet in point $E.$ Let $S$ be the degree-sum of angles $CDE$ and $DCE$, and let $S'$ represent the degree-sum of angles $BAD$ and $ABC.$ If $r=S/S'$, then:

$\text{(A) } r=1 \text{ sometimes, } r>1 \text{ sometimes}\quad\\ \text{(B) } r=1 \text{ sometimes, } r<1 \text{ sometimes}\quad\\ \text{(C) } 0<r<1\quad \text{(D) } r>1\quad \text{(E) } r=1$

Solution

Because $ABCD$ is a convex quadrilateral, the sum of its interior angles is $360^{\circ}$. Thus, $S'=\measuredangle BAD + \measuredangle ABC=360^{\circ}-(\measuredangle BCD+\measuredangle ADC)$. Furthermore, because $\angle BCD$ and $\angle ADC$ are supplementary to $\angle DCE$ and $\angle CDE$, respectively, the four angles sum to $2*180^{\circ}=360^{\circ}$, so $\measuredangle BCD+\measuredangle ADC=360^{\circ}-(\measuredangle DCE+\measuredangle CDE)=360^{\circ}-S$. Plussing this expression for $\measuredangle BCD+\measuredangle ADC$ into the first equation, we see that $S'=360^{\circ}-(360^{\circ}-S)=S$, so $r=\frac{S}{S'}=1$, which is answer choice $\fbox{E}$.

See also

1968 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1968_AHSME_Problems/Problem_6&oldid=222904"