1975 AHSME Problems/Problem 28
Contents
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
See Also
1975 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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