Difference between revisions of "1964 AHSME Problems/Problem 17"
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Using vector addition can help solve this problem quickly. Note that algebraically, adding <math>\overrightarrow{OP}</math> to <math>\overrightarrow{OQ}</math> will give <math>\overrightarrow{OR}</math>. One method of vector addition is literally known as the "parallelogram rule" - if you are given <math>\overrightarrow{OP}</math> and <math>\overrightarrow{OQ}</math>, to find <math>\overrightarrow{OR}</math>, you can literally draw a parallelogram, making a line though <math>P</math> parallel to <math>OQ</math>, and a line through <math>Q</math> parallel to <math>OP</math>. The intersection of those lines will give the fourth point <math>R</math>, and that fourth point will form a parallelogram with <math>O, P, Q</math>. | Using vector addition can help solve this problem quickly. Note that algebraically, adding <math>\overrightarrow{OP}</math> to <math>\overrightarrow{OQ}</math> will give <math>\overrightarrow{OR}</math>. One method of vector addition is literally known as the "parallelogram rule" - if you are given <math>\overrightarrow{OP}</math> and <math>\overrightarrow{OQ}</math>, to find <math>\overrightarrow{OR}</math>, you can literally draw a parallelogram, making a line though <math>P</math> parallel to <math>OQ</math>, and a line through <math>Q</math> parallel to <math>OP</math>. The intersection of those lines will give the fourth point <math>R</math>, and that fourth point will form a parallelogram with <math>O, P, Q</math>. |
Latest revision as of 20:38, 31 May 2020
Problem 17
Given the distinct points and . Line segments are drawn connecting these points to each other and to the origin . Of the three possibilities: (1) parallelogram (2) straight line (3) trapezoid, figure , depending upon the location of the points , and , can be:
Solution
Using vector addition can help solve this problem quickly. Note that algebraically, adding to will give . One method of vector addition is literally known as the "parallelogram rule" - if you are given and , to find , you can literally draw a parallelogram, making a line though parallel to , and a line through parallel to . The intersection of those lines will give the fourth point , and that fourth point will form a parallelogram with .
Thus, is a possibility. Case is also a possibility, if are collinear, then is also on that line.
Since and , which can be seen from either the prior reasoning or by examining slopes, the figure can never be a trapezoid, which requires exactly one of parallel sides.
Thus, the answer is
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
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