Difference between revisions of "1975 AHSME Problems/Problem 28"
(→Solution 2) |
(→Solution 3) |
||
Line 55: | Line 55: | ||
==Solution 3== | ==Solution 3== | ||
In order to find <math>\frac{EG}{FG}</math>, we can apply the law of sine to this model. Let: | In order to find <math>\frac{EG}{FG}</math>, we can apply the law of sine to this model. Let: | ||
− | + | ||
− | \angle MAC=\alpha, \; \angle MAB=\beta, \; \angle AGE=\gamma,\; \angle AMC=\delta | + | <math>\angle MAC=\alpha, \; \angle MAB=\beta, \; \angle AGE=\gamma,\; \angle AMC=\delta</math> |
− | + | ||
Then, in the <math>\triangle AMC</math> and <math>\triangle AMB</math>: | Then, in the <math>\triangle AMC</math> and <math>\triangle AMB</math>: | ||
− | + | ||
− | \frac{MC}{sin\alpha} | + | <math>\frac{MC}{sin\alpha}=\frac{AC}{sin\delta};</math> |
− | \frac{MB}{sin\beta} | + | <math>\frac{MB}{sin\beta}=\frac{AB}{sin\delta}=\frac{MC}{sin\alpha}\cdot\frac{3}{4};</math> |
− | \frac{sin\alpha}{sin\beta} | + | <math>\frac{sin\alpha}{sin\beta}=\frac{3}{4}</math> |
− | + | ||
In the <math>\triangle AGE</math> and <math>\triangle AGF</math>: | In the <math>\triangle AGE</math> and <math>\triangle AGF</math>: | ||
− | + | ||
− | \frac{FG}{sin\beta} | + | <math>\frac{FG}{sin\beta}=\frac{AF}{sin\gamma};</math> |
− | \frac{EG}{sin\alpha} | + | <math>\frac{EG}{sin\alpha}=\frac{AE}{sin\gamma};</math> |
− | \frac{2FG}{sin\beta} | + | <math>\frac{2FG}{sin\beta}=\frac{EG}{sin\alpha};</math> |
− | \frac{EG}{FG}=\frac{2sin\alpha}{sin\beta}=\frac{3}{2} | + | <math>\frac{EG}{FG}=\frac{2sin\alpha}{sin\beta}=\frac{3}{2}</math> |
− | + | ||
− | Hence, our answer is | + | Hence, our answer is \textbf{A}. |
-VSN | -VSN |
Revision as of 23:49, 11 May 2023
Problem 28
In shown in the adjoining figure, is the midpoint of side and . Points and are taken on and , respectively, and lines and intersect at . If then equals
Solution
Here, we use Mass Points. Let . We then have , , and Let have a mass of . Since is the midpoint, also has a mass of . Looking at segment , we have So Looking at segment ,we have So From this, we get and We want the value of . This can be written as Thus
~JustinLee2017
Solution 2
Since we only care about a ratio , and since we are given being the midpoint of , we realize we can conveniently also choose to be the midpoint of . (we're free to choose any point on as long as is twice , the constraint given in the problem). This means , and . We then connect which creates similar triangles and by SAS, and thus generates parallel lines and . This also immediately gives us similar triangles (note that because is in ratio).
~afroromanian
Solution 3
In order to find , we can apply the law of sine to this model. Let:
Then, in the and :
In the and :
Hence, our answer is \textbf{A}.
-VSN
See Also
1975 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.