Difference between revisions of "1968 AHSME Problems/Problem 6"
(Created page with "== Problem == Let side <math>AD</math> of convex quadrilateral <math>ABCD</math> be extended through <math>D</math>, and let side <math>BC</math> be extended through <math>C</ma...") |
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== Solution == | == 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 == | == 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]] | [[Category: Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:06, 17 July 2024
Problem
Let side of convex quadrilateral
be extended through
, and let side
be extended through
, to meet in point
Let
be the degree-sum of angles
and
, and let
represent the degree-sum of angles
and
If
, then:
Solution
Because is a convex quadrilateral, the sum of its interior angles is
. Thus,
. Furthermore, because
and
are supplementary to
and
, respectively, the four angles sum to
, so
. Plussing this expression for
into the first equation, we see that
, so
, which is answer choice
.
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 |
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