Difference between revisions of "1957 AHSME Problems/Problem 50"
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− | <math>\fbox{\textbf{(B)}remains stationary}</math>. | + | Let <math>AG=AA'=a</math> and <math>BG=BB'=b</math>. Then, we know that the diameter <math>d</math> of the circle equals <math>AG+BG=a+b</math>. Thus, because the circle's diameter does not change, <math>a+b</math> is constant. |
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+ | Because <math>A'O'=O'B'</math> and <math>AO=OB</math>, <math>\overline{OO'} \parallel \overline{AA'}</math>. Thus, <math>\overline{OO'} \perp \overline{AB}</math>, and so <math>OO'</math> is the distance from <math>O'</math> to <math>\overline{AB}</math>. | ||
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+ | Let <math>P</math> be some point which moves along <math>\overline{A'B'}</math>. Because <math>A'B'</math> is a line segment, as <math>P</math> moves from <math>A'</math> to <math>B'</math>, its distance from <math>\overline{AB}</math> will vary linearly with how much it has travelled along <math>\overline{A'B'}</math>. Thus, when it is halfway along <math>\overline{A'B'}</math> (in other words, when <math>P=O'</math>) its distance from <math>\overline{AB}</math> will be the arithmetic mean of its distance from <math>\overline{AB}</math> at <math>A'</math> (namely, <math>A'A=a</math>) and its distance from <math>\overline{AB}</math> at <math>B'</math> (namely, <math>B'B=b</math>). Thus, <math>O'O = \tfrac{a+b}2</math>. | ||
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+ | Because <math>a+b=d</math>, a constant, <math>\tfrac{a+b}2</math> is a constant as well. Thus, <math>OO'</math> is the same regardless of the position of <math>G</math>. Furthermore, from our work in paragraph 2, we know that <math>O'</math> must lie on the line perpendicular to <math>\overline{AB}</math> through point <math>O</math>. Therefore, because <math>O'</math> is a fixed distance from a fixed point on a fixed line, and it will not suddenly "jump across" to the other side of <math>\overline{AB}</math>, we can say with confidence that point <math>O'</math> <math>\fbox{\textbf{(B)} remains stationary}</math>. | ||
== See Also == | == See Also == |
Latest revision as of 16:18, 27 July 2024
Problem
In circle , is a moving point on diameter . is drawn perpendicular to and equal to . is drawn perpendicular to , on the same side of diameter as , and equal to . Let be the midpoint of . Then, as moves from to , point :
Solution
Let and . Then, we know that the diameter of the circle equals . Thus, because the circle's diameter does not change, is constant.
Because and , . Thus, , and so is the distance from to .
Let be some point which moves along . Because is a line segment, as moves from to , its distance from will vary linearly with how much it has travelled along . Thus, when it is halfway along (in other words, when ) its distance from will be the arithmetic mean of its distance from at (namely, ) and its distance from at (namely, ). Thus, .
Because , a constant, is a constant as well. Thus, is the same regardless of the position of . Furthermore, from our work in paragraph 2, we know that must lie on the line perpendicular to through point . Therefore, because is a fixed distance from a fixed point on a fixed line, and it will not suddenly "jump across" to the other side of , we can say with confidence that point .
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 49 |
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