Difference between revisions of "1957 AHSME Problems/Problem 47"
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== Solution == | == Solution == | ||
| − | <math>\boxed{\textbf{(A) }r\sqrt2}</math>. | + | Because <math>\overline{XY}</math> is a diameter of the circle and <math>\overline{AB} \perp \overline{XY}</math>, we know that <math>\overline{XY}</math> bisects <math>\overline{AB}</math>, so <math>AQ=BQ</math>. Thus, <math>M</math> is on the [[Perpendicular bisector#Locus|perpendicular bisector]] of <math>\overline{AB}</math>, and so <math>AM=BM</math>. Furthermore, by [[Thales' Theorem]], <math>\measuredangle AMB = 90^{\circ}</math>. Thus, because <math>\triangle AMB</math> is a right isosceles triangle with <math>AM=BM</math>, it is a [[45-45-90 triangle]]. Thus, <math>\measuredangle ABM = 45^{\circ}</math>. Now, draw <math>\overline{OA}</math> and <math>\overline{OD}</math>. Because <math>\angle ABD \cong \angle ABM</math> is an [[inscribed angle]] which intercepts minor arc <math>AD</math>, the measure of [[central angle]] <math>\angle AOD</math> must be <math>2\measuredangle ABD = 2 \cdot 45^{\circ} = 90^{\circ}</math>. Because <math>OA=OD=r</math>, <math>\triangle AOD</math> is also a <math>45</math>-<math>45</math>-<math>90</math> triangle, so <math>AD=\boxed{\textbf{(A) }r\sqrt2}</math>. |
== See Also == | == See Also == | ||
Latest revision as of 12:40, 27 July 2024
Problem
In circle 
, the midpoint of radius 
is 
; at , 
. The semi-circle with 
as diameter intersects 
in 
. Line 
intersects circle in 
, and line 
intersects circle in 
. Line 
is drawn. Then, if the radius of circle is 
, 
is:
![[asy] defaultpen(linewidth(.8pt)); unitsize(2.5cm); real m = 0; real b = 0; pair O = origin; pair X = (-1,0); pair Y = (1,0); pair Q = midpoint(O--X); pair A = (Q.x, -1*sqrt(3)/2); pair B = (Q.x, -1*A.y); pair M = (Q.x + sqrt(3)/2,0); m = (B.y - M.y)/(B.x - M.x); b = (B.y - m*B.x); pair D = intersectionpoint(Circle(O,1),M--(1.5,1.5*m + b)); m = (A.y - M.y)/(A.x - M.x); b = (A.y - m*A.x); pair C = intersectionpoint(Circle(O,1),M--(1.5,1.5*m + b)); draw(Circle(O,1)); draw(Arc(Q,sqrt(3)/2,-90,90)); draw(A--B); draw(X--Y); draw(B--D); draw(A--C); draw(A--D); dot(O);dot(M); label("$B$",B,NW); label("$C$",C,NE); label("$Y$",Y,E); label("$D$",D,SE); label("$A$",A,SW); label("$X$",X,W); label("$Q$",Q,SW); label("$O$",O,SW); label("$M$",M,NE+2N);[/asy]](http://latex.artofproblemsolving.com/d/4/0/d40adb845b830ab1e2f4f3b2600588d2140efaf7.png)
Solution
Because is a diameter of the circle and , we know that bisects , so 
. Thus, is on the perpendicular bisector of , and so 
. Furthermore, by Thales' Theorem, 
. Thus, because 
is a right isosceles triangle with , it is a 45-45-90 triangle. Thus, 
. Now, draw 
and 
. Because 
is an inscribed angle which intercepts minor arc , the measure of central angle 
must be 
. Because 
, 
is also a 
--
triangle, so 
.
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 46 |
Followed by Problem 48 | |
| 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. 
