Difference between revisions of "2017 AMC 10B Problems/Problem 8"
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==Solution 3== | ==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> | 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> | ||
| + | |||
| + | ==Solution 4== | ||
| + | |||
| + | As stated in solution 1, the triangle is isosceles. | ||
| + | |||
| + | <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> | ||
| + | |||
| + | This means that <math>D(-1, 3)</math> is the midpoint of <math>B(2, -3)</math> and <math>C(x,y)</math>. So <math>\frac{x+2}{2}</math> <math>= -1</math> and so <math>x = -4</math>. Similarly for <math>y</math>, we have <math>\frac{y-3}{2}</math> <math>=3</math> and so <math>y=9</math>. So our final answer is <math>\boxed{\textbf{(C) } (-4,9)}</math>. | ||
| + | |||
| + | - abed_nadir | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 12:38, 11 November 2023
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 
.
![[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]](http://latex.artofproblemsolving.com/d/6/c/d6c903419860d4d2d6fb832a27d73e8e7acd40ae.png)
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 . ~
Solution 4
As stated in solution 1, the triangle is isosceles.
This means that is the midpoint of and 
. So 
and so 
. Similarly for 
, we have 
and so 
. So our final answer is .
- abed_nadir
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. 
