Difference between revisions of "1957 AHSME Problems/Problem 26"
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== Solution == | == Solution == | ||
| + | |||
| + | <asy> | ||
| + | |||
| + | import geometry; | ||
| + | |||
| + | point A = (0,0); | ||
| + | point B = (3,1); | ||
| + | point C = (2,4); | ||
| + | point P = (A+B+C)/3; | ||
| + | |||
| + | draw(A--B--C--A); | ||
| + | dot(A); | ||
| + | label("A",A,SW); | ||
| + | dot(B); | ||
| + | label("B",B,SE); | ||
| + | dot(C); | ||
| + | label("C",C,NW); | ||
| + | dot(P); | ||
| + | label("P",P,E); | ||
| + | |||
| + | draw(A--P); | ||
| + | draw(B--P); | ||
| + | draw(C--P); | ||
| + | |||
| + | </asy> | ||
| + | |||
<math>\fbox{\textbf{(E) }the intersection of the medians of the triangle}</math>. | <math>\fbox{\textbf{(E) }the intersection of the medians of the triangle}</math>. | ||
Revision as of 15:42, 25 July 2024
Problem
From a point within a triangle, line segments are drawn to the vertices. A necessary and sufficient condition that the three triangles thus formed have equal areas is that the point be:
Solution
![[asy] import geometry; point A = (0,0); point B = (3,1); point C = (2,4); point P = (A+B+C)/3; draw(A--B--C--A); dot(A); label("A",A,SW); dot(B); label("B",B,SE); dot(C); label("C",C,NW); dot(P); label("P",P,E); draw(A--P); draw(B--P); draw(C--P); [/asy]](http://latex.artofproblemsolving.com/d/1/b/d1b0f48aa108bcb2f91149ab8f46ef8ba4fd5cb1.png)
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See Also
| 1957 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. 
