Difference between revisions of "1959 AHSME Problems/Problem 26"
Clarkculus (talk | contribs) (Created page with "== Problem 26== The base of an isosceles triangle is <math>\sqrt 2</math>. The medians to the leg intersect each other at right angles. The area of the triangle is: <math>\tex...") |
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Latest revision as of 12:26, 21 July 2024
Problem 26
The base of an isosceles triangle is 
. The medians to the leg intersect each other at right angles. The area of the triangle is:
Solution
Drop the altitude 
from the vertex 
to base 
, and note that it intersects the centroid 
. Then 
is an isosceles right triangle, so 
, so 
. By centroid properties, 
. Then the area of 
is 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 25 |
Followed by Problem 27 | |
| 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. 
