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Difference between revisions of "2025 SSMO Speed Round Problems/Problem 2"

(→‎Solution)
m (→‎Solution)
 
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A=(0,0);
 
A=(0,0);
 
B=(50,0);
 
B=(50,0);
−
M=(35,0);
+
N=(35,0);
−
N=(15,0);
+
M=(15,0);
  
 
draw(A--B);
 
draw(A--B);
Line 25: Line 25:
 
</asy>
 
</asy>
  
−
Note that <math>AN+BM = AB - MN = 30</math>. By AM-GM, the value of <math>AN\cdot BM</math> is at a maximum when <math>AN = BM = 15</math>. Thus, <math>AM = AN+MN = \boxed{35}</math>.
+
Note that <math>AN+BM = AB + MN = 70</math>. By AM-GM, the value of <math>AN\cdot BM</math> is at a maximum when <math>AN = BM = 35</math>. Thus, <math>AM = AN-MN = \boxed{15}</math>.
  
 
~Sedro
 
~Sedro

Latest revision as of 15:02, 9 September 2025

Problem

Let $A$ and $B$ be points such that $AB = 50$. Points $M$ and $N$ lie on $\overline{AB}$ such that $M$ lies between points $A$ and $N$ and $N$ lies between points $B$ and $M$. Given that $MN = 20$ and $AN \cdot BM$ is maximized, find the length of $AM$.

Solution

[asy] pair A,B,M,N; A=(0,0); B=(50,0); N=(35,0); M=(15,0); draw(A--B); dot(A,linewidth(4)); dot(B,linewidth(4)); dot(M,linewidth(4)); dot(N,linewidth(4)); label("$A$",A,dir(90)); label("$B$",B,dir(90)); label("$M$",M,dir(90)); label("$N$",N,dir(90)); [/asy]

Note that $AN+BM = AB + MN = 70$. By AM-GM, the value of $AN\cdot BM$ is at a maximum when $AN = BM = 35$. Thus, $AM = AN-MN = \boxed{15}$.

~Sedro

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