Difference between revisions of "1975 AHSME Problems/Problem 20"

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==Problem==
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In the adjoining figure triangle <math>ABC</math> is such that <math>AB = 4</math> and <math>AC = 8</math>. IF <math>M</math> is the midpoint of <math>BC</math> and <math>AM = 3</math>, what is the length of <math>BC</math>?
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<asy>
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draw((-4,0)--(4,0)--(-1,4)--cycle);
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draw((-1, 4)--(0, 0.00001));
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label("B", (-4,0), S);
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label("C", (4,0), S);
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label("A", (-1, 4), N);
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label("M", (0, 0.0001), S);
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</asy>
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==Solution==
 
Let <math>BM=CM=x</math>. Then, by Stewart's Theorem, we have
 
Let <math>BM=CM=x</math>. Then, by Stewart's Theorem, we have
 
<cmath>2x^3+18x=16x+64x</cmath>
 
<cmath>2x^3+18x=16x+64x</cmath>
 
<cmath>\implies x^2+9=40</cmath>
 
<cmath>\implies x^2+9=40</cmath>
 
<cmath>\implies x=\sqrt{31}\implies BC=\boxed{2\sqrt{31}}.</cmath>
 
<cmath>\implies x=\sqrt{31}\implies BC=\boxed{2\sqrt{31}}.</cmath>
The answer is <math>\textbf{(B)}.</math>
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The answer is <math>\boxed{B}.</math>
 
-brainiacmaniac31
 
-brainiacmaniac31
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==See Also==
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{{AHSME box|year=1975|num-b=19|num-a=21}}
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{{MAA Notice}}

Revision as of 16:29, 19 January 2021

Problem

In the adjoining figure triangle $ABC$ is such that $AB = 4$ and $AC = 8$. IF $M$ is the midpoint of $BC$ and $AM = 3$, what is the length of $BC$?

[asy] draw((-4,0)--(4,0)--(-1,4)--cycle); draw((-1, 4)--(0, 0.00001)); label("B", (-4,0), S); label("C", (4,0), S); label("A", (-1, 4), N); label("M", (0, 0.0001), S); [/asy]

Solution

Let $BM=CM=x$. Then, by Stewart's Theorem, we have \[2x^3+18x=16x+64x\] \[\implies x^2+9=40\] \[\implies x=\sqrt{31}\implies BC=\boxed{2\sqrt{31}}.\] The answer is $\boxed{B}.$ -brainiacmaniac31

See Also

1975 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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