Difference between revisions of "2017 AMC 10B Problems/Problem 8"
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draw(A--B--C--cycle); | draw(A--B--C--cycle); | ||
draw(A--D); | draw(A--D); | ||
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label("$A$",A,E); | label("$A$",A,E); | ||
label("$B$",B,S); | label("$B$",B,S); |
Latest revision as of 16:14, 13 September 2024
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 . ~
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 .
- youtube.com/indianmathguy
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.