Difference between revisions of "1970 AHSME Problems/Problem 30"
m (→Problem) |
(Added solution) |
||
| Line 16: | Line 16: | ||
== Solution == | == Solution == | ||
| − | <math>\ | + | With reference to the diagram above, let <math>E</math> be the point on <math>AB</math> such that <math>DE||BC</math>. Let <math>\angle ABC=\alpha</math>. We then have <math>\alpha =\angle AED = \angle EDC</math> since <math>AB||CD</math>, so <math>\angle ADE=\angle ADC-\angle BDC=2\alpha-\alpha = \alpha</math>, which means <math>\triangle AED</math> is isosceles. |
| + | Therefore, <math>AB=AE+EB=a+b</math>, hence our answer is <math>\fbox{E}</math>. | ||
== See also == | == See also == | ||
{{AHSME 35p box|year=1970|num-b=29|num-a=31}} | {{AHSME 35p box|year=1970|num-b=29|num-a=31}} | ||
Latest revision as of 07:57, 28 June 2018
Problem
![[asy] draw((0,0)--(2,2)--(5/2,1/2)--(2,0)--cycle,dot); MP("A",(0,0),W);MP("B",(2,2),N);MP("C",(5/2,1/2),SE);MP("D",(2,0),S); MP("a",(1,0),N);MP("b",(17/8,1/8),N); [/asy]](http://latex.artofproblemsolving.com/4/2/9/429211325ffed98c6590a7895780ee9afb3a9f0b.png)
In the accompanying figure, segments 
and 
are parallel, the measure of angle 
is twice that of angle 
, and the measures of segments 
and are 
and 
respectively. Then the measure of is equal to
Solution
With reference to the diagram above, let 
be the point on such that 
. Let 
. We then have 
since 
, so 
, which means 
is isosceles.
Therefore, 
, hence our answer is 
.
See also
| 1970 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Problem 31 | |
| 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. 
