Difference between revisions of "1981 AHSME Problems/Problem 19"
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== Solution == | == Solution == | ||
+ | |||
+ | <asy> | ||
+ | size(150); | ||
+ | defaultpen(linewidth(0.7)+fontsize(10)); | ||
+ | pair B=origin, A=14*dir(42), C=intersectionpoint(B--(30,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b), Q=2*N-B; | ||
+ | draw(N--B--A--N--M--C--A^^B--M); | ||
+ | draw(N--Q); | ||
+ | markscalefactor=0.1; | ||
+ | draw(rightanglemark(B,N,A)); | ||
+ | markscalefactor=0.3; | ||
+ | draw(anglemark(B,A,N)); | ||
+ | markscalefactor=0.2; | ||
+ | draw(anglemark(N,A,Q)); | ||
+ | markscalefactor=0.1; | ||
+ | pair point=N; | ||
+ | label("$A$", A, dir(point--A)); | ||
+ | label("$B$", B, dir(point--B)); | ||
+ | label("$C$", C, dir(point--C)); | ||
+ | label("$M$", M, dir(-45)); | ||
+ | label("$N$", N, dir(45)); | ||
+ | label("$Q$", Q, dir(45)); | ||
+ | |||
+ | label(rotate(angle(dir(A--Q)))*"$14$", A--Q, dir(A--Q)*dir(90)); | ||
+ | label(rotate(angle(dir(Q--C)))*"$5$", Q--C, dir(Q--C)*dir(90)); | ||
+ | label(rotate(angle(dir(A--B)))*"$14$", A--B, dir(A--B)*dir(90)); | ||
+ | </asy> | ||
+ | |||
Extend <math>BN</math> to meet <math>AC</math> at <math>Q</math>. Then <math>\triangle BNM \sim \triangle BQC</math>, so <math>BN=NQ</math> and <math>QC=19-AQ=2MN</math>. | Extend <math>BN</math> to meet <math>AC</math> at <math>Q</math>. Then <math>\triangle BNM \sim \triangle BQC</math>, so <math>BN=NQ</math> and <math>QC=19-AQ=2MN</math>. | ||
Latest revision as of 21:00, 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.