Difference between revisions of "2016 AMC 10B Problems/Problem 9"
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==Solution== | ==Solution== | ||
− | + | <asy>import graph;size(7cm,IgnoreAspect); | |
− | + | real f(real x) {return x*x;} | |
− | + | draw((0,0)--(4,16)--(-4,16)--cycle,blue); | |
− | + | draw(graph(f,-5,5,operator ..),gray); | |
+ | xaxis("$x$");yaxis("$y$",-1); | ||
+ | label("$y=x^2$",(4.5,20.25),E); | ||
+ | draw((4.2,0)--(4.2,16),Arrows); | ||
+ | label("$r^2$",(4.2,0)--(4.2,16),E); | ||
+ | draw((0,17)--(4,17),Arrows); | ||
+ | label("$r$",(0,17)--(4,17),N); | ||
+ | </asy> | ||
+ | The area of the triangle is <math>\frac{(2r)(r^2)}{2} = r^3</math>, so <math>r^3=64\implies r=4</math>, giving a total distance across the top of <math>8</math>, which is answer <math>\textbf{(C)}</math>. | ||
+ | |||
+ | ==Solution 2 (Guess and Check)== | ||
+ | |||
+ | Let the point where the height of the triangle intersects with the base be <math>D</math>. Now we can guess what <math>x</math> is and find <math>y</math>. If <math>x</math> is <math>3</math>, then <math>y</math> is <math>9</math>. The cords of <math>B</math> and <math>C</math> would be <math>(-3,9)</math> and <math>(3,9)</math>, respectively. The distance between <math>B</math> and <math>C</math> is <math>6</math>, meaning the area would be <math>\frac{6 \cdot 9}{2}=27</math>, not <math>64</math>. Now we let <math>x=4</math>. <math>y</math> would be <math>16</math>. The cords of <math>B</math> and <math>C</math> would be <math>(-4,16)</math> and <math>(4,16)</math>, respectively. <math>BC</math> would be <math>8</math>, and the height would be <math>16</math>. The area would then be <math>\frac{8 \cdot 16}{2}</math> which is <math>64</math>, so <math>BC</math> is <math>\boxed{\textbf{(C)}\ 8}</math>. | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/pSJkO6kQGOs | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/1pi0eiD3jHc | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=8|num-a=10}} | {{AMC10 box|year=2016|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:45, 2 July 2023
Contents
Problem
All three vertices of lie on the parabola defined by
, with
at the origin and
parallel to the
-axis. The area of the triangle is
. What is the length of
?
Solution
The area of the triangle is
, so
, giving a total distance across the top of
, which is answer
.
Solution 2 (Guess and Check)
Let the point where the height of the triangle intersects with the base be . Now we can guess what
is and find
. If
is
, then
is
. The cords of
and
would be
and
, respectively. The distance between
and
is
, meaning the area would be
, not
. Now we let
.
would be
. The cords of
and
would be
and
, respectively.
would be
, and the height would be
. The area would then be
which is
, so
is
.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.