Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2000 AMC 8 Problems/Problem 24"

(→‎Video Solution)
(→‎Problem)
(6 intermediate revisions by 3 users not shown)
Line 21: Line 21:
 
label("$G$",G,NW);</asy>
 
label("$G$",G,NW);</asy>
  
−
<math> \text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\circ\qquad\text{(E)}\ 90^\circ </math>
+
$ \text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\
  
−
==Solution==
+
== Video Solution ==
−
 
+
https://www.youtube.com/watch?v=8ntXubG2Iho  ~David
−
As a strategy, think of how <math>\angle B + \angle D</math> would be determined, particularly without determining either of the angles individually, since it may not be possible to determine <math>\angle B</math> or <math>\angle D</math> alone.  If you see <math>\triangle BFD</math>, the you can see that the problem is solved quickly after determining <math>\angle BFD</math>.
 
−
 
 
−
But start with <math>\triangle AGF</math>, since that's where most of our information is.  Looking at <math>\triangle AGF</math>, since <math>\angle F = \angle G</math>, and <math>\angle A = 20</math>, we can write:
 
−
 
 
−
<math>\angle A + \angle G + \angle F = 180</math>
 
−
 
 
−
<math>20 + 2\angle F = 180</math>
 
−
 
 
−
<math>\angle AFG = 80</math>
 
−
 
 
−
By noting that <math>\angle AFG</math> and <math>\angle GFD</math> make a straight line, we know
 
−
 
 
−
<math>\angle AFG + \angle GFD = 180</math>
 
−
 
 
−
<math>80 + \angle GFD = 180</math>
 
−
 
 
−
<math>\angle GFD = 100</math>
 
−
 
 
−
Ignoring all other parts of the figure and looking only at <math>\triangle BFD</math>, you see that <math>\angle B + \angle D + \angle F = 180</math>. But <math>\angle F</math> is the same as <math>\angle GFD</math>. Therefore:
 
−
 
 
−
<math>\angle B + \angle D + \angle GFD = 180</math>
 
−
<math>\angle B + \angle D + 100 = 180</math>
 
−
<math>\angle B + \angle D = 80^\circ</math>, and the answer is thus <math>\boxed{D}</math>
 
  
 
==See Also==
 
==See Also==

Revision as of 19:56, 1 September 2024

Problem

If $\angle A = 20^\circ$ and $\angle AFG =\angle AGF$, then $\angle B+\angle D =$

[asy] pair A,B,C,D,EE,F,G; A = (0,0); B = (9,4); C = (21,0); D = (13,-12); EE = (4,-16); F = (13/2,-6); G = (8,0); draw(A--C--EE--B--D--cycle); label("$A$",A,W); label("$B$",B,N); label("$C$",C,E); label("$D$",D,SE); label("$E$",EE,SW); label("$F$",F,WSW); label("$G$",G,NW);[/asy]

$ \text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\

Video Solution

https://www.youtube.com/watch?v=8ntXubG2Iho ~David

See Also

2000 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_24&oldid=226168"