1964 AHSME Problems/Problem 39
Problem
The magnitudes of the sides of triangle 
are 
, 
, and 
, as shown, with 
. Through interior point 
and the vertices 
, 
, 
, lines are drawn meeting the opposite sides in 
, 
, 
, respectively. Let 
. Then, for all positions of point , 
is less than:
![[asy] import math; pair A = (0,0), B = (1,3), C = (5,0), P = (1.5,1); pair X = extension(B,C,A,P), Y = extension(A,C,B,P), Z = extension(A,B,C,P); draw(A--B--C--cycle); draw(A--X); draw(B--Y); draw(C--Z); dot(P); dot(A); dot(B); dot(C); label("$A$",A,dir(210)); label("$B$",B,dir(90)); label("$C$",C,dir(-30)); label("$A'$",X,dir(-100)); label("$B'$",Y,dir(65)); label("$C'$",Z,dir(20)); label("$P$",P,dir(70)); label("$a$",X,dir(80)); label("$b$",Y,dir(-90)); label("$c$",Z,dir(110)); [/asy]](http://latex.artofproblemsolving.com/b/8/1/b8175dd073e666965c2dd3700b5b55a1823d0505.png)
Solution
We know that in a 
, if 
then 
, we can use this fact in the different triangles to form inequalities, and then add the inequalities.
In 
, since 
, we have 
by the above argument.
Now, 
, hence we have 
And, 
, hence we have 
And, 
, hence we have 
Finally, adding all three inequalities, we have $b + a + a > AX + BY + ZC \imples AX + BY + CZ < 2a + b$ (Error compiling LaTeX. Unknown error_msg)
See Also
| 1964 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 38 |
Followed by Problem 40 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
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