Difference between revisions of "1961 AHSME Problems/Problem 40"
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By Cauchy-Schwarz, | By Cauchy-Schwarz, | ||
− | <cmath>(5^2 + 12^2)(x^2 + y^2) | + | <cmath>(5^2 + 12^2)(x^2 + y^2) >= (5x+12y)^2</cmath> |
Therefore: | Therefore: | ||
<cmath>(13^2)(x^2 + y^2) \geq 60^2</cmath> | <cmath>(13^2)(x^2 + y^2) \geq 60^2</cmath> | ||
Line 54: | Line 54: | ||
Therefore: | Therefore: | ||
<cmath>\sqrt{x^2 + y^2} \geq \frac{60}{13}</cmath> | <cmath>\sqrt{x^2 + y^2} \geq \frac{60}{13}</cmath> | ||
+ | Thus the answer is <math>\boxed{\textbf{(A)}}</math>. | ||
==See Also== | ==See Also== |
Revision as of 21:30, 11 June 2020
Problem
Find the minimum value of if .
Solutions
Solution 1
Let , so . Thus, this problem is really finding the shortest distance from the origin to the line .
From the graph, the shortest distance from the origin to the line is the altitude to the hypotenuse of the right triangle with legs and . The hypotenuse is and the area is , so the altitude to the hypotenuse is , which is answer choice .
Solution 2
Solve for in the linear equation. Substitute in . To find the minimum, find the vertex of the quadratic. The x-value of the vertex is . Thus, the minimum value is The answer is .
Solution 3
By Cauchy-Schwarz, Therefore: Therefore: Thus the answer is .
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 39 |
Followed by Last Question | |
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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