Difference between revisions of "1993 AHSME Problems/Problem 27"
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Start by considering the triangle traced by <math>P</math> as the circle moves around the triangle. It turns out this triangle is similar to the <math>6-8-10</math> triangle (Proof: Realize that the slope of the line made while the circle is on <math>AC</math> is the same as line <math>AC</math> and that it makes a right angle when the circle switches from being on <math>AB</math> to <math>BC</math>). Then, drop the perpendiculars as shown. | Start by considering the triangle traced by <math>P</math> as the circle moves around the triangle. It turns out this triangle is similar to the <math>6-8-10</math> triangle (Proof: Realize that the slope of the line made while the circle is on <math>AC</math> is the same as line <math>AC</math> and that it makes a right angle when the circle switches from being on <math>AB</math> to <math>BC</math>). Then, drop the perpendiculars as shown. | ||
− | Since the smaller triangle is also a <math>6-8-10 = 3-4-5</math> triangle, we can label the sides <math>EF, CE, </math>and <math>DF</math> as <math>3x, 4x,</math> and <math>5x</math> respectively. Now, it is clear that <math>GB = DE + 1 = 4x + 1</math>, so <math>AH = AG = 8 - GB = 7 - 4x</math> since <math>AH</math> and <math>AG</math> are both tangent to the circle P at some point. We can apply the same logic to the other side as well to get <math>CI = 5 - 3x</math>. Finally, since we have <math>HI = DF = 5x</math>, we have <math>AC = 10 = (7 - 4x) + (5x) + (5 - 3x) = 12 - 2x</math>, so <math>x = 1</math> and <math>3x + 4x + 5x = \fbox{(B) 12}</math> | + | Since the smaller triangle is also a <math>6-8-10 = 3-4-5</math> triangle, we can label the sides <math>EF,</math> <math>CE, </math>and <math>DF</math> as <math>3x, 4x,</math> and <math>5x</math> respectively. Now, it is clear that <math>GB = DE + 1 = 4x + 1</math>, so <math>AH = AG = 8 - GB = 7 - 4x</math> since <math>AH</math> and <math>AG</math> are both tangent to the circle P at some point. We can apply the same logic to the other side as well to get <math>CI = 5 - 3x</math>. Finally, since we have <math>HI = DF = 5x</math>, we have <math>AC = 10 = (7 - 4x) + (5x) + (5 - 3x) = 12 - 2x</math>, so <math>x = 1</math> and <math>3x + 4x + 5x = \fbox{(B) 12}</math> |
-Solution by Someonenumber011 | -Solution by Someonenumber011 |
Revision as of 10:09, 18 July 2020
Problem
The sides of have lengths and . A circle with center and radius rolls around the inside of , always remaining tangent to at least one side of the triangle. When first returns to its original position, through what distance has traveled?
Solution
Start by considering the triangle traced by as the circle moves around the triangle. It turns out this triangle is similar to the triangle (Proof: Realize that the slope of the line made while the circle is on is the same as line and that it makes a right angle when the circle switches from being on to ). Then, drop the perpendiculars as shown.
Since the smaller triangle is also a triangle, we can label the sides and as and respectively. Now, it is clear that , so since and are both tangent to the circle P at some point. We can apply the same logic to the other side as well to get . Finally, since we have , we have , so and
-Solution by Someonenumber011
See also
1993 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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 |
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