Difference between revisions of "1975 AHSME Problems/Problem 20"
(Created page with "Let <math>BM=CM=x</math>. Then, by Stewart's Theorem, we have <cmath>2x^3+18x=16x+64x</cmath> <cmath>\implies x^2+9=40</cmath> <cmath>\implies x=\sqrt{31}\implies BC=\boxed{2\...") |
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| + | ==Problem== | ||
| + | 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>? | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <math> | ||
| + | \textbf{(A)}\ 2\sqrt{26} \qquad \textbf{(B)}\ 2\sqrt{31} \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 4 + 2\sqrt{13}\\ \textbf{(E)}\ \text{not enough information given to solve the problem} | ||
| + | </math> | ||
| + | |||
| + | ==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>\boxed{B}.</math> | ||
-brainiacmaniac31 | -brainiacmaniac31 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1975|num-b=19|num-a=21}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:32, 19 January 2021
Problem
In the adjoining figure triangle 
is such that 
and 
. IF 
is the midpoint of 
and 
, what is the length of ?
![[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]](http://latex.artofproblemsolving.com/f/4/2/f42fe5450102486dda1fdec4c1520e55ac6dee3e.png)
Solution
Let 
. Then, by Stewart's Theorem, we have
![\[2x^3+18x=16x+64x\]](http://latex.artofproblemsolving.com/a/f/d/afd3cf1a1a368ef79905d4e8d3fa41029f467bc9.png)
![\[\implies x^2+9=40\]](http://latex.artofproblemsolving.com/e/3/9/e391428c480e31e2bb9e9ff00e008e3893142584.png)
![\[\implies x=\sqrt{31}\implies BC=\boxed{2\sqrt{31}}.\]](http://latex.artofproblemsolving.com/1/8/1/181584bcc40552dafcf9b933095297d4ac21c89c.png)
The answer is 
-brainiacmaniac31
See Also
| 1975 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
