Difference between revisions of "1970 AHSME Problems/Problem 15"
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The trisection points of <math>(-4, 5)</math> and <math>(5, -1)</math> can be found by trisecting the x-coordinates and the y-coordinates separately. The difference of the x-coordinates is <math>9</math>, so the trisection points happen at <math>-4 + \frac{9}{3}</math> and <math>-4 + \frac{9}{3} + \frac{9}{3}</math>, which are <math>-1</math> and <math>2</math>. Similarly, the y-coordinates have a difference of <math>6</math>, so the trisections happen at <math>3</math> and <math>1</math>. So, the two points are <math>(-1, 3)</math> and <math>(2, 1)</math>. | The trisection points of <math>(-4, 5)</math> and <math>(5, -1)</math> can be found by trisecting the x-coordinates and the y-coordinates separately. The difference of the x-coordinates is <math>9</math>, so the trisection points happen at <math>-4 + \frac{9}{3}</math> and <math>-4 + \frac{9}{3} + \frac{9}{3}</math>, which are <math>-1</math> and <math>2</math>. Similarly, the y-coordinates have a difference of <math>6</math>, so the trisections happen at <math>3</math> and <math>1</math>. So, the two points are <math>(-1, 3)</math> and <math>(2, 1)</math>. | ||
− | We now check which line has both <math>(3, 4)</math> and one of the two trisection points on it. Plugging in <math>(x, y) = (3, 4)</math> into all five of the equations works. The point <math>(2, 1)</math> doesn't work in any of the five lines. However, (-1, 3)<math> works in line < | + | We now check which line has both <math>(3, 4)</math> and one of the two trisection points on it. Plugging in <math>(x, y) = (3, 4)</math> into all five of the equations works. The point <math>(2, 1)</math> doesn't work in any of the five lines. However, <math>(-1, 3)</math> works in line <math>E</math>. Thus, the answer is <math>\fbox{E}</math> |
== See also == | == See also == |
Latest revision as of 21:26, 13 July 2019
Problem
Lines in the -plane are drawn through the point and the trisection points of the line segment joining the points and . One of these lines has the equation
Solution
The trisection points of and can be found by trisecting the x-coordinates and the y-coordinates separately. The difference of the x-coordinates is , so the trisection points happen at and , which are and . Similarly, the y-coordinates have a difference of , so the trisections happen at and . So, the two points are and .
We now check which line has both and one of the two trisection points on it. Plugging in into all five of the equations works. The point doesn't work in any of the five lines. However, works in line . Thus, the answer is
See also
1970 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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All AHSME Problems and Solutions |
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