Difference between revisions of "1959 AHSME Problems/Problem 32"
m (fixed typo) |
(→Solution) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 4: | Line 4: | ||
== Solution == | == Solution == | ||
| − | |||
| + | <asy> | ||
| + | |||
| + | import geometry; | ||
| + | |||
| + | point O=(0,0); | ||
| + | point A=(5,0); | ||
| + | point B,T; | ||
| + | |||
| + | circle c=circle(O,3); | ||
| + | |||
| + | markscalefactor=0.05; | ||
| + | |||
| + | // Circle, segment OA | ||
| + | draw(c); | ||
| + | dot(O); | ||
| + | label("O",O,NW); | ||
| + | dot(A); | ||
| + | label("A",A,NE); | ||
| + | draw(O--A); | ||
| + | |||
| + | // Segments OT, OA | ||
| + | line[] t1=tangents(c,A); | ||
| + | pair[] t=intersectionpoints(t1[0], c); | ||
| + | T=t[0]; | ||
| + | dot(t[0]); | ||
| + | label("T",T,SE); | ||
| + | draw(A--T--O); | ||
| + | draw(rightanglemark(A,T,O)); | ||
| + | |||
| + | // Point B | ||
| + | pair[] b=intersectionpoints((O--A),c); | ||
| + | B=b[0]; | ||
| + | dot("B",B,NE); | ||
| + | |||
| + | // Length labels | ||
| + | label("$r$",midpoint(O--T),SW); | ||
| + | label("$r$",midpoint(O--B),N); | ||
| + | label("$\frac{4}{3}r$",midpoint(A--T),SE); | ||
| + | |||
| + | </asy> | ||
| + | |||
| + | Let the circle have center <math>O</math>, let the point of tangency be point <math>T</math>, and let <math>B</math> be the intersection of <math>\overline{OA}</math> with the circle, as in the diagram. By the definitions of a circle and a tangent to a circle, we know that <math>OB=OT=r</math> and <math>\overline{OT} \perp \overline{TA}</math>. By the [[Pythagorean Theorem]], <math>OA=\sqrt{\frac{25r^2}{9}}=\frac{5}{3}r</math>. Because the shortest segment from an external point to a circle lies on the line connecting that point to the center of the circle, our desired distance is <math>AB</math>. Because <math>OA=\frac{5}{3}r</math> and <math>OB=r</math>, <math>AB=\frac{5}{3}r-r=\frac{2}{3}r=\frac{(4/3)r}{2}=\boxed{\textbf{(C) }\frac{1}{2}l}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 21:15, 20 July 2024
Problem
The length 
of a tangent, drawn from a point 
to a circle, is 
of the radius 
. The (shortest) distance from A to the circle is:
Solution
![[asy] import geometry; point O=(0,0); point A=(5,0); point B,T; circle c=circle(O,3); markscalefactor=0.05; // Circle, segment OA draw(c); dot(O); label("O",O,NW); dot(A); label("A",A,NE); draw(O--A); // Segments OT, OA line[] t1=tangents(c,A); pair[] t=intersectionpoints(t1[0], c); T=t[0]; dot(t[0]); label("T",T,SE); draw(A--T--O); draw(rightanglemark(A,T,O)); // Point B pair[] b=intersectionpoints((O--A),c); B=b[0]; dot("B",B,NE); // Length labels label("$r$",midpoint(O--T),SW); label("$r$",midpoint(O--B),N); label("$\frac{4}{3}r$",midpoint(A--T),SE); [/asy]](http://latex.artofproblemsolving.com/b/0/a/b0a7d663479391704b13abf3c42068b309b73d7f.png)
Let the circle have center 
, let the point of tangency be point 
, and let 
be the intersection of 
with the circle, as in the diagram. By the definitions of a circle and a tangent to a circle, we know that 
and 
. By the Pythagorean Theorem, 
. Because the shortest segment from an external point to a circle lies on the line connecting that point to the center of the circle, our desired distance is 
. Because 
and 
, 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 31 |
Followed by Problem 33 | |
| 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. 
