1961 AHSME Problems/Problem 36: Difference between revisions
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) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 1: | Line 1: | ||
== Problem | == Problem == | ||
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</math>, find the length of <math>AB</math>. | 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</math>, find the length of <math>AB</math>. | ||
| Line 7: | Line 7: | ||
\textbf{(C)}\ 4.25\qquad | \textbf{(C)}\ 4.25\qquad | ||
\textbf{(D)}\ 2\sqrt{5} \qquad | \textbf{(D)}\ 2\sqrt{5} \qquad | ||
\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== | ||
Latest revision as of 10:15, 2 June 2018
Problem
In 
the median from 
is given perpendicular to the median from 
. If 
and 
, find the length of 
.
Solution
![[asy] draw((-16,0)--(8,0)); draw((-16,0)--(16,-24)); draw((16,-24)--(0,24)--(0,-12)); draw((-16,0)--(0,24)); draw((0,2)--(2,2)--(2,0)); draw((0,-12)--(8,0),dotted); dot((16,-24)); label("C",(16,-24),SE); dot((-16,0)); label("A",(-16,0),W); dot((0,24)); label("B",(0,24),N); label("3",(8,-18),SW); label("3",(-8,-6),SW); label("3.5",(12,-12),NE); label("3.5",(4,12),NE); dot((0,-12)); label("M",(0,-12),SW); dot((8,0)); label("N",(8,0),NE); dot((0,0)); label("G",(0,0),NW); [/asy]](http://latex.artofproblemsolving.com/1/9/a/19a40f9471bcdf5b25ca03b8c795b99ada390be9.png)
By SAS Similarity, 
, so 
. Thus, by AA Similarity, 
.
Let 
and 
, so 
and 
. By the Pythagorean Theorem,
![\[4a^2 + b^2 = 9\]](http://latex.artofproblemsolving.com/8/4/c/84c96a210e140418e57cdead11a4a0c56fc88cc9.png)
![\[a^2 + 4b^2 = \frac{49}{4}\]](http://latex.artofproblemsolving.com/1/c/b/1cb9e70c0d5beb8a7c7cea240f72a867fd8472c1.png)
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
