Difference between revisions of "1957 AHSME Problems/Problem 50"
(created solution page) |
(→Solution) |
||
(One intermediate revision by the same user not shown) | |||
Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
− | <math>\fbox{\textbf{(B)}remains stationary}</math>. | + | |
+ | <asy> | ||
+ | |||
+ | import geometry; | ||
+ | |||
+ | point A = (0,0); | ||
+ | point B = (16,0); | ||
+ | point O = midpoint(A--B); | ||
+ | point G = (11,0); | ||
+ | point A1, B1, O1; | ||
+ | |||
+ | // Defining A', B', and O' | ||
+ | pair[] a1 = intersectionpoints(circle(A,length(segment(A,G))),perpendicular(A,line(A,B))); | ||
+ | A1 = a1[1]; | ||
+ | pair[] b1 = intersectionpoints(circle(B, length(segment(B,G))),perpendicular(B,line(A,B))); | ||
+ | B1 = b1[1]; | ||
+ | O1 = midpoint(A1--B1); | ||
+ | |||
+ | // Circle w/ Diameter | ||
+ | draw(circle(O,length(segment(A,B))/2)); | ||
+ | draw(A--B); | ||
+ | |||
+ | // Segments AA', BB', A'B', and OO' | ||
+ | draw(A--A1); | ||
+ | draw(B--B1); | ||
+ | draw(A1--B1); | ||
+ | draw(O--O1); | ||
+ | |||
+ | // Points w/ Labels | ||
+ | dot(A); | ||
+ | label("A",A,SW); | ||
+ | dot(B); | ||
+ | label("B",B,SE); | ||
+ | dot(A1); | ||
+ | label("A$'$",A1,NW); | ||
+ | dot(B1); | ||
+ | label("B$'$",B1,NE); | ||
+ | dot(O); | ||
+ | label("O",O,S); | ||
+ | dot(O1); | ||
+ | label("O$'$",O1,N); | ||
+ | dot(G); | ||
+ | label("G",G,S); | ||
+ | |||
+ | // Right angle marks | ||
+ | markscalefactor = 0.11; | ||
+ | draw(rightanglemark(A1,A,B)); | ||
+ | draw(rightanglemark(B1,B,A)); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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 |
Followed by Last Problem | |
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.