Difference between revisions of "2017 AMC 10B Problems/Problem 8"
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<math>\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)</math> | <math>\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)</math> | ||
+ | |||
==Solution 1== | ==Solution 1== | ||
Since <math>AB = AC</math>, then <math>\triangle ABC</math> is isosceles, so <math>BD = CD</math>. Therefore, the coordinates of <math>C</math> are <math>(-1 - 3, 3 + 6) = \boxed{\textbf{(C) } (-4,9)}</math>. | Since <math>AB = AC</math>, then <math>\triangle ABC</math> is isosceles, so <math>BD = CD</math>. Therefore, the coordinates of <math>C</math> are <math>(-1 - 3, 3 + 6) = \boxed{\textbf{(C) } (-4,9)}</math>. | ||
+ | |||
+ | <asy> | ||
+ | pair A,B,C,D; | ||
+ | A=(11,9); | ||
+ | B=(2,-3); | ||
+ | C=(-4,9); | ||
+ | D=(-1,3); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(A--D); | ||
+ | draw(rightanglemark(A,D,B)); | ||
+ | label("$A$",A,E); | ||
+ | label("$B$",B,S); | ||
+ | label("$D$",D,W); | ||
+ | label("$C$",C,N); | ||
+ | </asy> | ||
==Solution 2== | ==Solution 2== | ||
Calculating the equation of the line running between points <math>B</math> and <math>D</math>, <math>y = -2x + 1</math>. The only coordinate of <math>C</math> that is also on this line is <math>\boxed{\textbf{(C) } (-4,9)}</math>. | Calculating the equation of the line running between points <math>B</math> and <math>D</math>, <math>y = -2x + 1</math>. The only coordinate of <math>C</math> that is also on this line is <math>\boxed{\textbf{(C) } (-4,9)}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Similar to the first solution, because the triangle is isosceles, then the line drawn in the middle separates the triangle into two smaller congruent triangles. To get from <math>B</math> to <math>D</math>, we go to the right <math>3</math> and up <math>6</math>. Then to get to point <math>C</math> from point <math>D</math>, we go to the right <math>3</math> and up <math>6</math>, getting us the coordinates <math>\boxed{\textbf{(C) } (-4,9)}</math>. ~<math>\text{KLBBC}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/4rRckA3gcPU | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/XRfOULUmWbY?t=367 | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=7|num-a=9}} | {{AMC10 box|year=2017|ab=B|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:17, 25 July 2021
Contents
Problem
Points and are vertices of with . The altitude from meets the opposite side at . What are the coordinates of point ?
Solution 1
Since , then is isosceles, so . Therefore, the coordinates of are .
Solution 2
Calculating the equation of the line running between points and , . The only coordinate of that is also on this line is .
Solution 3
Similar to the first solution, because the triangle is isosceles, then the line drawn in the middle separates the triangle into two smaller congruent triangles. To get from to , we go to the right and up . Then to get to point from point , we go to the right and up , getting us the coordinates . ~
Video Solution
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/XRfOULUmWbY?t=367
~IceMatrix
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.