Difference between revisions of "1952 AHSME Problems/Problem 49"
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| − | {{AHSME 50p box|year=1952|num-b= | + | {{AHSME 50p box|year=1952|num-b=48|num-a=50}} |
| − | [[Category: | + | [[Category: Intermediate Geometry Problems]] |
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:57, 2 October 2014
Problem
![[asy] unitsize(27); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A,B,C,D,E,F,X,Y,Z; A=(3,3); B=(0,0); C=(6,0); D=(4,0); E=(4,2); F=(1,1); draw(A--B--C--cycle); draw(A--D); draw(B--E); draw(C--F); X=intersectionpoint(A--D,C--F); Y=intersectionpoint(B--E,A--D); Z=intersectionpoint(B--E,C--F); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S); label("$E$",E,NE); label("$F$",F,NW); label("$N_1$",X,NE); label("$N_2$",Y,WNW); label("$N_3$",Z,S); [/asy]](http://latex.artofproblemsolving.com/c/1/d/c1d737a4088cc4d9061c296f96c1fa7232d8d8d0.png)
In the figure, 
, 
and 
are one-third of their respective sides. It follows that 
, and similarly for lines BE and CF. Then the area of triangle 
is:
Solution
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 48 |
Followed by Problem 50 | |
| 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. 
