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Difference between revisions of "1964 AHSME Problems/Problem 39"

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Now, <math>\angle AXC > \angle B \ge \angle C</math>, hence we have <math>AC > AX \implies b > AX</math>
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Now, <math>\angle AA'C > \angle B \ge \angle C</math>, hence we have <math>AC > AA' \implies b > AA'</math>
  
−
And, <math>\angle BYC > \angle A \ge \angle C</math>, hence we have <math>BY > BC \implies a > BY</math>
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And, <math>\angle BB'C > \angle A \ge \angle C</math>, hence we have <math>BC > BB' \implies a > BB'</math>
  
−
And, <math>\angle CZB > \angle A \ge \angle B</math>, hence we have <math>BC > ZC \implies a > ZC</math>
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And, <math>\angle CC'B > \angle A \ge \angle B</math>, hence we have <math>BC > CC' \implies a > CC'</math>
  
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Finally, adding all three inequalities, we have <math>b + a + a > AX + BY + ZC \implies AX + BY + CZ < \textbf{2a + b}</math>
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Finally, adding all three inequalities, we have <math>b + a + a > AA' + BB' + CC' \implies AA' + BB' + CC' < \textbf{2a + b}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 21:34, 7 March 2023

Problem

The magnitudes of the sides of triangle $ABC$ are $a$, $b$, and $c$, as shown, with $c\le b\le a$. Through interior point $P$ and the vertices $A$, $B$, $C$, lines are drawn meeting the opposite sides in $A'$, $B'$, $C'$, respectively. Let $s=AA'+BB'+CC'$. Then, for all positions of point $P$, $s$ is less than:

$\textbf{(A) }2a+b\qquad\textbf{(B) }2a+c\qquad\textbf{(C) }2b+c\qquad\textbf{(D) }a+2b\qquad \textbf{(E) }$ $a+b+c$


[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]

Solution

We know that in a $\triangle DEF$, if $\angle D \le \angle E$ then $EF \le DF$, we can use this fact in the different triangles to form inequalities, and then add the inequalities.

In $\triangle ABC$, since $c \le b \le a$, we have $\angle C \le \angle B \le \angle A$ by the above argument.


Now, $\angle AA'C > \angle B \ge \angle C$, hence we have $AC > AA' \implies b > AA'$

And, $\angle BB'C > \angle A \ge \angle C$, hence we have $BC > BB' \implies a > BB'$

And, $\angle CC'B > \angle A \ge \angle B$, hence we have $BC > CC' \implies a > CC'$

Finally, adding all three inequalities, we have $b + a + a > AA' + BB' + CC' \implies AA' + BB' + CC' < \textbf{2a + b}$

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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