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Difference between revisions of "1964 AHSME Problems/Problem 36"

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==Problem==
 
In this figure the radius of the circle is equal to the altitude of the equilateral triangle <math>ABC</math>. The circle is made to roll along the side <math>AB</math>, remaining tangent to it at a variable point <math>T</math> and intersecting lines <math>AC</math> and <math>BC</math> in variable points <math>M</math> and <math>N</math>, respectively. Let <math>n</math> be the number of degrees in arc <math>MTN</math>. Then <math>n</math>, for all permissible positions of the circle:
 
In this figure the radius of the circle is equal to the altitude of the equilateral triangle <math>ABC</math>. The circle is made to roll along the side <math>AB</math>, remaining tangent to it at a variable point <math>T</math> and intersecting lines <math>AC</math> and <math>BC</math> in variable points <math>M</math> and <math>N</math>, respectively. Let <math>n</math> be the number of degrees in arc <math>MTN</math>. Then <math>n</math>, for all permissible positions of the circle:
  
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</asy>
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==Solution==
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E
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==See Also==
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{{AHSME 40p box|year=1964|num-b=35|num-a=37}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 19:28, 18 April 2020

Problem

In this figure the radius of the circle is equal to the altitude of the equilateral triangle $ABC$. The circle is made to roll along the side $AB$, remaining tangent to it at a variable point $T$ and intersecting lines $AC$ and $BC$ in variable points $M$ and $N$, respectively. Let $n$ be the number of degrees in arc $MTN$. Then $n$, for all permissible positions of the circle:

$\textbf{(A) }\text{varies from }30^{\circ}\text{ to }90^{\circ}$

$\textbf{(B) }\text{varies from }30^{\circ}\text{ to }60^{\circ}$

$\textbf{(C) }\text{varies from }60^{\circ}\text{ to }90^{\circ}$

$\textbf{(D) }\text{remains constant at }30^{\circ}$

$\textbf{(E) }\text{remains constant at }60^{\circ}$

[asy] pair A = (0,0), B = (1,0), C = dir(60), T = (2/3,0); pair M = intersectionpoint(A--C,Circle((2/3,sqrt(3)/2),sqrt(3)/2)), N = intersectionpoint(B--C,Circle((2/3,sqrt(3)/2),sqrt(3)/2)); draw((0,0)--(1,0)--dir(60)--cycle); draw(Circle((2/3,sqrt(3)/2),sqrt(3)/2)); label("$A$",A,dir(210)); label("$B$",B,dir(-30)); label("$C$",C,dir(90)); label("$M$",M,dir(190)); label("$N$",N,dir(75)); label("$T$",T,dir(-90)); [/asy]

Solution

E

See Also

1964 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 35 		Followed by
Problem 37
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 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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