Difference between revisions of "1961 AHSME Problems/Problem 36"
Rockmanex3 (talk | contribs) (Created page with "== Problem 36== In <math>\triangle ABC</math> the median from <math>A</math> is given perpendicular to the median from <math>B</math>. If <math>BC=7</math> and <math>AC=6</ma...") |
Rockmanex3 (talk | contribs) (Solution to Problem 36) |
||
Line 9: | Line 9: | ||
\textbf{(E)}\ 4.5 </math> | \textbf{(E)}\ 4.5 </math> | ||
− | ==Solution | + | ==Solution== |
<asy> | <asy> | ||
Line 39: | Line 39: | ||
</asy> | </asy> | ||
− | By | + | By SAS Similarity, <math>\triangle ABC \sim \triangle MNC</math>, so <math>AB \parallel MN</math>. Thus, by AA Similarity, <math>\triangle AGB \sim \triangle NGM</math>. |
Let <math>a = GN</math> and <math>b = GM</math>, so <math>AG = 2a</math> and <math>BG = 2b</math>. By the [[Pythagorean Theorem]], | Let <math>a = GN</math> and <math>b = GM</math>, so <math>AG = 2a</math> and <math>BG = 2b</math>. By the [[Pythagorean Theorem]], | ||
<cmath>4a^2 + b^2 = 9</cmath> | <cmath>4a^2 + b^2 = 9</cmath> | ||
<cmath>a^2 + 4b^2 = \frac{49}{4}</cmath> | <cmath>a^2 + 4b^2 = \frac{49}{4}</cmath> | ||
+ | Adding the two equations yields <math>5a^2 + 5b^2 = \frac{85}{4}</math>, so <math>a^2 + b^2 = \frac{17}{4}</math>. Thus, <math>MN = \frac{\sqrt{17}}{2}</math>, so <math>AB = \sqrt{17}</math>, which is answer choice <math>\boxed{\textbf{(B)}}</math>. | ||
==See Also== | ==See Also== |
Revision as of 11:12, 2 June 2018
Problem 36
In the median from is given perpendicular to the median from . If and , find the length of .
Solution
By SAS Similarity, , so . Thus, by AA Similarity, .
Let and , so and . By the Pythagorean Theorem, Adding the two equations yields , so . Thus, , so , which is answer choice .
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 35 |
Followed by Problem 37 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.