Difference between revisions of "1982 AHSME Problems/Problem 14"
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==Solution== | ==Solution== | ||
| − | Drop a perpendicular from <math>N</math> to <math>AG</math> at point <math>H</math>. <math>AN=45</math>, and since <math>\triangle{AGP}</math> is similar to <math>\triangle{AHN}</math>. <math>NH=9</math>. <math>NE=NF=15</math> so by the Pythagorean Theorem, <math>EH=HF=12</math>. Thus <math>EF=\boxed{24.}</math> Answer is then <math>\boxed{C}</math>. | + | Drop a perpendicular line from <math>N</math> to <math>AG</math> at point <math>H</math>. <math>AN=45</math>, and since <math>\triangle{AGP}</math> is similar to <math>\triangle{AHN}</math>. <math>NH=9</math>. <math>NE=NF=15</math> so by the Pythagorean Theorem, <math>EH=HF=12</math>. Thus <math>EF=\boxed{24.}</math> Answer is then <math>\boxed{C}</math>. |
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1982|num-b=13|num-a=15}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 23:29, 29 June 2025
Problem 14
In the adjoining figure, points 
and 
lie on line segment 
, and 
, and 
are diameters of circle 
, and 
, respectively. Circles , and all have radius 
and the line 
is tangent to circle at 
. If intersects circle 
at points 
and 
, then chord 
has length
![[asy] size(250); defaultpen(fontsize(10)); pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1]; draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P); label("$A$", A, W); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, dir(0)); label("$P$", P, S); label("$N$", N, S); label("$O$", O, S); label("$E$", E, dir(120)); label("$F$", F, NE); label("$G$", G, dir(100));[/asy]](http://latex.artofproblemsolving.com/3/4/d/34d6e023825b451845bd167422e5cd1c057e76ba.png)
Solution
Drop a perpendicular line from to at point 
. 
, and since 
is similar to 
. 
. 
so by the Pythagorean Theorem, 
. Thus 
Answer is then 
.
See Also
| 1982 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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. 
