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

(Created page with "==Problem== Let <math>A</math> and <math>B</math> be points such that <math>AB = 50</math>. Points <math>M</math> and <math>N</math> lie on <math>\overline{AB}</math> such th...")
 
(→‎Solution)
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
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<asy>
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pair A,B,M,N;
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A=(0,0);
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B=(50,0);
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M=(35,0);
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N=(15,0);
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draw(A--B);
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dot(A,linewidth(4));
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dot(B,linewidth(4));
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dot(M,linewidth(4));
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dot(N,linewidth(4));
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label("$A$",A,dir(90));
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label("$B$",B,dir(90));
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label("$M$",M,dir(90));
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label("$N$",N,dir(90));
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</asy>
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 +
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>.
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~Sedro

Revision as of 14:50, 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); M=(35,0); N=(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 $AN+BM = AB - MN = 30$. By AM-GM, the value of $AN\cdot BM$ is at a maximum when $AN = BM = 15$. Thus, $AM = AN+MN = \boxed{35}$.

~Sedro

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