Difference between revisions of "1951 AHSME Problems/Problem 46"
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==Problem== | ==Problem== | ||
| + | <math>AB</math> is a fixed diameter of a circle whose center is <math>O</math>. From <math>C</math>, any point on the circle, a chord <math>CD</math> is drawn perpendicular to <math>AB</math>. Then, as <math>C</math> moves over a semicircle, the bisector of angle <math>OCD</math> cuts the circle in a point that always: | ||
| + | |||
| + | <math> \textbf{(A)}\ \text{bisects the arc }AB\qquad\textbf{(B)}\ \text{trisects the arc }AB\qquad\textbf{(C)}\ \text{varies} </math> | ||
| + | <math> \textbf{(D)}\ \text{is as far from }AB\text{ as from }D\qquad\textbf{(E)}\ \text{is equidistant from }B\text{ and }C </math> | ||
| + | |||
==Solution== | ==Solution== | ||
| − | + | <asy> | |
| + | pair O=(0,0), A=(-1,0), B=(1,0), C=(-0.5,0.5sqrt(3)), | ||
| + | D=(-0.5,-0.5sqrt(3)), E=(0.5,-0.5sqrt(3)), P=(0,-1); | ||
| + | draw(circle(O,1)); | ||
| + | label("$A$",A,W); label("$B$",B,E); label("$O$",O,NE); label("$P$",P,S); | ||
| + | label("$C$",C,NW); label("$D$",D,SW); label ("$E$",E,SE); | ||
| + | dot(A); dot(B); dot(C); dot(D); dot(E); dot(O); dot(P); | ||
| + | draw(A--B); draw(C--D--E--cycle); draw(C--P,dashed); | ||
| + | </asy> | ||
| + | Draw in diameter <math>CE</math>. Note that, as it is inscribed in a semicircle, <math> \bigtriangleup CDE</math> always has a right angle at <math>D</math>. Since it is given that <math> AB \perp CD </math>, <math> AB \parallel DE </math> by congruent corresponding angles. <math> CP </math> bisects <math> DE </math>, as well as the arc subtended by <math> DE </math>, <math> \widehat{DPE} </math>. Because <math> AB \parallel DE </math>, <math> CP </math> always <math> \boxed{\textbf{(A)}\ \text{bisects the arc }AB} </math>. | ||
== See Also == | == See Also == | ||
{{AHSME 50p box|year=1951|num-b=45|num-a=47}} | {{AHSME 50p box|year=1951|num-b=45|num-a=47}} | ||
Latest revision as of 16:26, 1 January 2014
Problem
is a fixed diameter of a circle whose center is 
. From 
, any point on the circle, a chord 
is drawn perpendicular to . Then, as moves over a semicircle, the bisector of angle 
cuts the circle in a point that always:
Solution
![[asy] pair O=(0,0), A=(-1,0), B=(1,0), C=(-0.5,0.5sqrt(3)), D=(-0.5,-0.5sqrt(3)), E=(0.5,-0.5sqrt(3)), P=(0,-1); draw(circle(O,1)); label("$A$",A,W); label("$B$",B,E); label("$O$",O,NE); label("$P$",P,S); label("$C$",C,NW); label("$D$",D,SW); label ("$E$",E,SE); dot(A); dot(B); dot(C); dot(D); dot(E); dot(O); dot(P); draw(A--B); draw(C--D--E--cycle); draw(C--P,dashed); [/asy]](http://latex.artofproblemsolving.com/d/3/e/d3e0bfe83a71846b39ca5ae22dc534fd89efb34b.png)
Draw in diameter 
. Note that, as it is inscribed in a semicircle, 
always has a right angle at 
. Since it is given that 
, 
by congruent corresponding angles. 
bisects 
, as well as the arc subtended by , 
. Because , always 
.
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 45 |
Followed by Problem 47 | |
| 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. 
