Difference between revisions of "1995 AHSME Problems/Problem 9"
(New page: ==Problem== Consider the figure consisting of a square, its diagonals, and the segments joining the midpoints of opposite sides. The total number of triangles of any size in the figure is ...) |
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draw((0,1)--(2,1));</asy> | draw((0,1)--(2,1));</asy> | ||
| − | There are 8 little triangles, 4 triangles with twice the area, and 4 triangles with four times the area of the smaller triangles. <math>8+4+4=16\Rightarrow \mathrm{( | + | There are 8 little triangles, 4 triangles with twice the area, and 4 triangles with four times the area of the smaller triangles. <math>8+4+4=16\Rightarrow \mathrm{(D)}</math> |
==See also== | ==See also== | ||
| + | {{AHSME box|year=1995|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:59, 5 July 2013
Problem
Consider the figure consisting of a square, its diagonals, and the segments joining the midpoints of opposite sides. The total number of triangles of any size in the figure is
Solution
![[asy]draw((0,0)--(2,2)); draw((0,2)--(2,0)); draw((0,0)--(0,2)); draw((0,2)--(2,2)); draw((2,2)--(2,0)); draw((0,0)--(2,0)); draw((1,0)--(1,2)); draw((0,1)--(2,1));[/asy]](http://latex.artofproblemsolving.com/8/1/b/81bef27ccad078d2c07aad39603746ee913aeb0b.png)
There are 8 little triangles, 4 triangles with twice the area, and 4 triangles with four times the area of the smaller triangles. 
See also
| 1995 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
