Difference between revisions of "1981 AHSME Problems/Problem 19"
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Since <math>\angle ANB=90^\circ = \angle ANQ</math>, <math>\angle BAN=\angle NAQ</math> (since <math>AN</math> is an angle bisector) and <math>\triangle ANB</math> and <math>\triangle ANQ</math> share side <math>AN</math>, <math>\triangle ANB \cong \triangle ANQ</math>. Thus <math>AQ=14</math>, and so <math>MN=\frac{19-AQ}{2}=\frac{5}{2}</math>, hence our answer is <math>\fbox{B}</math>. | Since <math>\angle ANB=90^\circ = \angle ANQ</math>, <math>\angle BAN=\angle NAQ</math> (since <math>AN</math> is an angle bisector) and <math>\triangle ANB</math> and <math>\triangle ANQ</math> share side <math>AN</math>, <math>\triangle ANB \cong \triangle ANQ</math>. Thus <math>AQ=14</math>, and so <math>MN=\frac{19-AQ}{2}=\frac{5}{2}</math>, hence our answer is <math>\fbox{B}</math>. | ||
+ | |||
+ | ==See also== | ||
+ | |||
+ | {{AHSME box|year=1981|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} |
Revision as of 14:12, 28 June 2025
Problem 19
In ,
is the midpoint of side
,
bisects
,
, and
is the measure of
. If sides
and
have lengths
and
, respectively, then find
.
Solution
Extend to meet
at
. Then
, so
and
.
Since ,
(since
is an angle bisector) and
and
share side
,
. Thus
, and so
, hence our answer is
.
See also
1981 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 | ||
All AHSME Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.