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Difference between revisions of "1982 AHSME Problems/Problem 14"
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<math>\bold{1982 AHSME Problems/Problem 14}</math>
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==Problem 14==
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<math>\bold{Problem 14:}</math>
 
  
 
In the adjoining figure, points <math>B</math> and <math>C</math> lie on line segment <math>AD</math>, and <math>AB, BC</math>, and <math>CD</math> are diameters of circle  <math>O, N</math>, and <math>P</math>, respectively. Circles <math>O, N</math>, and <math>P</math> all have radius <math>15</math> and the line <math>AG</math> is tangent to circle <math>P</math> at <math>G</math>. If <math>AG</math> intersects circle <math>N</math> at points <math>E</math> and <math>F</math>, then chord <math>EF</math> has length
 
In the adjoining figure, points <math>B</math> and <math>C</math> lie on line segment <math>AD</math>, and <math>AB, BC</math>, and <math>CD</math> are diameters of circle  <math>O, N</math>, and <math>P</math>, respectively. Circles <math>O, N</math>, and <math>P</math> all have radius <math>15</math> and the line <math>AG</math> is tangent to circle <math>P</math> at <math>G</math>. If <math>AG</math> intersects circle <math>N</math> at points <math>E</math> and <math>F</math>, then chord <math>EF</math> has length
  
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[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("<math>A</math>", A, W); label("<math>B</math>", B, SE); label("<math>C</math>", C, NE); label("<math>D</math>", D, dir(0)); label("<math>P</math>", P, S); label("<math>N</math>", N, S); label("<math>O</math>", O, S); label("<math>E</math>", E, dir(120)); label("<math>F</math>", F, NE); label("<math>G</math>", G, dir(100));[/asy]
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<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>
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<math>\bold{Solution:}</math>
 
  
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Since <math>GP</math> is 15, <math>AP</math> is 75, and <math>\angle{AGP}=90</math>, <math>AG=15\sqrt{24}</math>.
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==Solution==
  
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Now drop an altitude 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 Pythagorean Theorem, <math>EH=HF=12</math>. Thus <math>EF=\boxed{24}</math>. <math>\boxed{C}</math>
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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>.

Latest revision as of 21:35, 16 August 2022

Problem 14

In the adjoining figure, points $B$ and $C$ lie on line segment $AD$, and $AB, BC$, and $CD$ are diameters of circle $O, N$, and $P$, respectively. Circles $O, N$, and $P$ all have radius $15$ and the line $AG$ is tangent to circle $P$ at $G$. If $AG$ intersects circle $N$ at points $E$ and $F$, then chord $EF$ 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]

Solution

Drop a perpendicular line from $N$ to $AG$ at point $H$. $AN=45$, and since $\triangle{AGP}$ is similar to $\triangle{AHN}$. $NH=9$. $NE=NF=15$ so by the Pythagorean Theorem, $EH=HF=12$. Thus $EF=\boxed{24.}$ Answer is then $\boxed{C}$.

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