Difference between revisions of "1995 AHSME Problems/Problem 17"
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Given regular pentagon <math>ABCDE</math>, a circle can be drawn that is tangent to <math>\overline{DC}</math> at <math>D</math> and to <math>\overline{AB}</math> at <math>A</math>. The number of degrees in minor arc <math>AD</math> is | Given regular pentagon <math>ABCDE</math>, a circle can be drawn that is tangent to <math>\overline{DC}</math> at <math>D</math> and to <math>\overline{AB}</math> at <math>A</math>. The number of degrees in minor arc <math>AD</math> is | ||
| − | [[ | + | <!-- original image at [[File:1995 AHSME num.17.png]] --> |
| + | <asy>size(100); defaultpen(linewidth(0.7)); | ||
| + | draw(rotate(18)*polygon(5)); | ||
| + | real x=0.6180339887; | ||
| + | draw(Circle((-x,0), 1)); | ||
| + | int i; | ||
| + | for(i=0; i<5; i=i+1) { | ||
| + | dot(origin+1*dir(36+72*i)); | ||
| + | } | ||
| + | |||
| + | label("$B$", origin+1*dir(36+72*0), dir(origin--origin+1*dir(36+72*0))); | ||
| + | label("$A$", origin+1*dir(36+72*1), dir(origin--origin+1*dir(36+72))); | ||
| + | label("$E$", origin+1*dir(36+72*2), dir(origin--origin+1*dir(36+144))); | ||
| + | label("$D$", origin+1*dir(36+72*3), dir(origin--origin+1*dir(36+72*3))); | ||
| + | label("$C$", origin+1*dir(36+72*4), dir(origin--origin+1*dir(36+72*4))); | ||
| + | </asy> | ||
<math> \mathrm{(A) \ 72 } \qquad \mathrm{(B) \ 108 } \qquad \mathrm{(C) \ 120 } \qquad \mathrm{(D) \ 135 } \qquad \mathrm{(E) \ 144 } </math> | <math> \mathrm{(A) \ 72 } \qquad \mathrm{(B) \ 108 } \qquad \mathrm{(C) \ 120 } \qquad \mathrm{(D) \ 135 } \qquad \mathrm{(E) \ 144 } </math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | Define major arc DA as DA, and minor arc DA as da. Extending DC and AB to meet at F, we see that <math>\angle CFB=36=\frac{DA-da}{2}</math>. We now have two equations: <math>DA-da=72</math>, and <math>DA+da=360</math>. Solving, <math>DA=216</math> and <math>da=144\Rightarrow \mathrm{(E)}</math>. | + | Define major arc DA as <math>DA</math>, and minor arc DA as <math>da</math>. Extending DC and AB to meet at F, we see that <math>\angle CFB=36=\frac{DA-da}{2}</math>. We now have two equations: <math>DA-da=72</math>, and <math>DA+da=360</math>. Solving, <math>DA=216</math> and <math>da=144\Rightarrow \mathrm{(E)}</math>. |
==See also== | ==See also== | ||
| Line 13: | Line 28: | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:10, 19 June 2024
Problem
Given regular pentagon 
, a circle can be drawn that is tangent to 
at 
and to 
at 
. The number of degrees in minor arc 
is
![[asy]size(100); defaultpen(linewidth(0.7)); draw(rotate(18)*polygon(5)); real x=0.6180339887; draw(Circle((-x,0), 1)); int i; for(i=0; i<5; i=i+1) { dot(origin+1*dir(36+72*i)); } label("$B$", origin+1*dir(36+72*0), dir(origin--origin+1*dir(36+72*0))); label("$A$", origin+1*dir(36+72*1), dir(origin--origin+1*dir(36+72))); label("$E$", origin+1*dir(36+72*2), dir(origin--origin+1*dir(36+144))); label("$D$", origin+1*dir(36+72*3), dir(origin--origin+1*dir(36+72*3))); label("$C$", origin+1*dir(36+72*4), dir(origin--origin+1*dir(36+72*4))); [/asy]](http://latex.artofproblemsolving.com/6/7/9/67955d58098fd50e5d4caa0fd57150a361ae17af.png)
Solution 1
Define major arc DA as 
, and minor arc DA as 
. Extending DC and AB to meet at F, we see that 
. We now have two equations: 
, and 
. Solving, 
and 
.
See also
| 1995 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
