Difference between revisions of "1996 AHSME Problems/Problem 20"
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3) <math>\widehat {BC}</math>, where <math>BC</math> is an arc around the circle. | 3) <math>\widehat {BC}</math>, where <math>BC</math> is an arc around the circle. | ||
− | The actual path will go <math>A \rightarrow B \rightarrow C \rightarrow D</math>, so the | + | The actual path will go <math>A \rightarrow B \rightarrow C \rightarrow D</math>, so the actual segments will be in order <math>1, 3, 2</math>. |
Let <math>O</math> be the center of the circle at <math>(6,8)</math>. | Let <math>O</math> be the center of the circle at <math>(6,8)</math>. | ||
Line 21: | Line 21: | ||
<math>OA = 10</math> and <math>OB = 5</math> since <math>B</math> is on the circle. Since <math>\triangle OAB</math> is a right triangle with right angle <math>B</math>, we find that <math>AB = \sqrt{10^2 - 5^2} = 5\sqrt{3}</math>. This means that <math>\triangle OAB</math> is a <math>30-60-90</math> triangle with sides <math>5:5\sqrt{3}:10</math>. | <math>OA = 10</math> and <math>OB = 5</math> since <math>B</math> is on the circle. Since <math>\triangle OAB</math> is a right triangle with right angle <math>B</math>, we find that <math>AB = \sqrt{10^2 - 5^2} = 5\sqrt{3}</math>. This means that <math>\triangle OAB</math> is a <math>30-60-90</math> triangle with sides <math>5:5\sqrt{3}:10</math>. | ||
− | + | Thus, <math>\angle DOC = 60^\circ</math>, and <math>\angle AOB = 60^\circ</math>. Since <math>\angle AOB</math>, <math>\angle DOC</math> and <math>\angle BOC</math> lie on a straight line, <math>\angle BOC</math> must be <math>60^\circ</math> as well. Thus, the arc that we travel is a <math>60^\circ</math> arc, and we travel <math>\frac{C}{6} = \frac{2\pi r}{6} = \frac{2\pi \cdot 5}{6} = \frac{5\pi}{3}</math> around the circle. | |
Thus, <math>AB = 5\sqrt{3}</math>, <math>\widehat {BC} = \frac{5\pi}{3}</math>, and <math>{CD} = 5\sqrt{3}</math>. The total distance is <math>10\sqrt{3} + \frac{5\pi}{3}</math>, which is option <math>\boxed{C}</math>. | Thus, <math>AB = 5\sqrt{3}</math>, <math>\widehat {BC} = \frac{5\pi}{3}</math>, and <math>{CD} = 5\sqrt{3}</math>. The total distance is <math>10\sqrt{3} + \frac{5\pi}{3}</math>, which is option <math>\boxed{C}</math>. | ||
+ | |||
+ | ==Solution 2 (by elimination of options)== | ||
+ | The shortest path likely consists of two congruent line segments connected by an arc of the circle, so the total length can be expressed in the form | ||
+ | <cmath>a + b\pi,</cmath> | ||
+ | where \(a\) and \(b\) are positive real numbers. | ||
+ | |||
+ | Using this observation, we can eliminate options A, B, and D, as they do not fit this form. | ||
+ | |||
+ | Next, the total circumference of the circle is \(10\pi\). Since half the circumference is \(5\pi\), option E can be ruled out because it corresponds to exactly half the circle's circumference, which is unlikely to be the shortest path involving line segments. | ||
+ | |||
+ | Therefore, by elimination of the other choices, the only remaining option is \(10\sqrt{3} + \frac{5\pi}{3}\), which matches option \(\boxed{C}\). | ||
+ | |||
+ | ~Voidling | ||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=19|num-a=21}} | {{AHSME box|year=1996|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:24, 9 August 2025
Problem 20
In the xy-plane, what is the length of the shortest path from to
that does not go inside the circle
?
Solution
The pathway from to
will consist of three segments:
1) , where
is tangent to the circle at point
.
2) , where
is tangent to the circle at point
.
3) , where
is an arc around the circle.
The actual path will go , so the actual segments will be in order
.
Let be the center of the circle at
.
and
since
is on the circle. Since
is a right triangle with right angle
, we find that
. This means that
is a
triangle with sides
.
Thus, , and
. Since
,
and
lie on a straight line,
must be
as well. Thus, the arc that we travel is a
arc, and we travel
around the circle.
Thus, ,
, and
. The total distance is
, which is option
.
Solution 2 (by elimination of options)
The shortest path likely consists of two congruent line segments connected by an arc of the circle, so the total length can be expressed in the form
where \(a\) and \(b\) are positive real numbers.
Using this observation, we can eliminate options A, B, and D, as they do not fit this form.
Next, the total circumference of the circle is \(10\pi\). Since half the circumference is \(5\pi\), option E can be ruled out because it corresponds to exactly half the circle's circumference, which is unlikely to be the shortest path involving line segments.
Therefore, by elimination of the other choices, the only remaining option is \(10\sqrt{3} + \frac{5\pi}{3}\), which matches option \(\boxed{C}\).
~Voidling
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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.