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Difference between revisions of "1970 AHSME Problems/Problem 30"

(Created page with "== Problem == In the accompanying figure, segments <math>AB</math> and <math>CD</math> are parallel, the measure of angle <math>D</math> is twice that of angle <math>B</math>, a...")
 
(Added solution)
 
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== Problem ==
 
== 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>
  
 
In the accompanying figure, segments <math>AB</math> and <math>CD</math> are parallel, the measure of angle <math>D</math> is twice that of angle <math>B</math>, and the measures of segments <math>AD</math> and <math>CD</math> are <math>a</math> and <math>b</math> respectively. Then the measure of <math>AB</math> is equal to
 
In the accompanying figure, segments <math>AB</math> and <math>CD</math> are parallel, the measure of angle <math>D</math> is twice that of angle <math>B</math>, and the measures of segments <math>AD</math> and <math>CD</math> are <math>a</math> and <math>b</math> respectively. Then the measure of <math>AB</math> is equal to
Line 10: Line 16:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{E}</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 ==
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{{AHSME 35p box|year=1970|num-b29|num-a=31}}   
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{{AHSME 35p box|year=1970|num-b=29|num-a=31}}   
  
 
[[Category: Intermediate Geometry Problems]]
 
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

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]

In the accompanying figure, segments $AB$ and $CD$ are parallel, the measure of angle $D$ is twice that of angle $B$, and the measures of segments $AD$ and $CD$ are $a$ and $b$ respectively. Then the measure of $AB$ is equal to

$\text{(A) } \tfrac{1}{2}a+2b\quad \text{(B) } \tfrac{3}{2}b+\tfrac{3}{4}a\quad \text{(C) } 2a-b\quad \text{(D) } 4b-\tfrac{1}{2}a\quad \text{(E) } a+b$

Solution

With reference to the diagram above, let $E$ be the point on $AB$ such that $DE||BC$. Let $\angle ABC=\alpha$. We then have $\alpha =\angle AED = \angle EDC$ since $AB||CD$, so $\angle ADE=\angle ADC-\angle BDC=2\alpha-\alpha = \alpha$, which means $\triangle AED$ is isosceles.

Therefore, $AB=AE+EB=a+b$, hence our answer is $\fbox{E}$.

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

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