Difference between revisions of "1968 AHSME Problems/Problem 6"

Latest revision as of 18:06, 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

$[asy] draw((-16,0)--(16,-24)); draw((0,24)--(16,-24)); draw((-16,0)--(0,24)); draw((0,-12)--(8,0)); dot((16,-24)); label("E",(16,-24),SE); dot((-16,0)); label("A",(-16,0),W); dot((0,24)); label("B",(0,24),N); dot((0,-12)); label("D",(0,-12),SW); dot((8,0)); label("C",(8,0),NE); [/asy]$

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 $\fbox{E}$ .

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.

Page

Toolbox

Search

Difference between revisions of "1968 AHSME Problems/Problem 6"

Latest revision as of 18:06, 17 July 2024

Problem

Solution

See also

@@ Line 10: / Line 10: @@
 == Solution ==
-<math>\fbox{}</math>
+<asy>
+draw((-16,0)--(16,-24));
+draw((0,24)--(16,-24));
+draw((-16,0)--(0,24));
+draw((0,-12)--(8,0));
+dot((16,-24));
+label("E",(16,-24),SE);
+dot((-16,0));
+label("A",(-16,0),W);
+dot((0,24));
+label("B",(0,24),N);
+dot((0,-12));
+label("D",(0,-12),SW);
+dot((8,0));
+label("C",(8,0),NE);
+</asy>
+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 ==
-{{AHSME box|year=1968|num-b=5|num-a=7}}
+{{AHSME 35p box|year=1968|num-b=5|num-a=7}}
 [[Category: Introductory Geometry Problems]]
 {{MAA Notice}}