Difference between revisions of "1999 AMC 8 Problems/Problem 23"
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− | ==Video Solution== | + | ==Solution 2== |
− | https://www.youtube.com/watch?v=AH5_Gol2GCM | + | Connect <math>AC</math>, <math>S_\triangle AMC=S_\triangle ANC</math>. To satisfied the three area is equal, we have <math>2S_\triangle AMC=S_\triangle BMC</math>, <math>2S_\triangle ANC=S_\triangle DNC</math>. Thus, <math>AM=AN=\frac{1}{2}BM=\frac{1}{2}AB=1</math>. <math>BM=2,BC=3,MC=\boxed{\text{(C)}\ \sqrt{13}}</math>. |
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+ | == Video Solution by CosineMethod [🔥Fast and Easy🔥]== | ||
+ | |||
+ | https://www.youtube.com/watch?v=AH5_Gol2GCM | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=1999|num-b=22|num-a=24}} | {{AMC8 box|year=1999|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 04:33, 12 March 2024
Contents
Problem
Square has sides of length 3. Segments and divide the square's area into three equal parts. How long is segment ?
Solution
Since the square has side length , the area of the entire square is .
The segments divide the square into 3 equal parts, so the area of each part is .
Since has area and base , using the area formula for a triangle:
Thus, height .
Since is a right triangle, .
Solution 2
Connect , . To satisfied the three area is equal, we have , . Thus, . .
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=AH5_Gol2GCM
See Also
1999 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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.