Difference between revisions of "1965 AHSME Problems/Problem 37"
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− | ==Solution== | + | == Problem == |
− | We use mass points for this problem. Let <math>\text{m} A</math> denote the mass of point <math>A</math>. | + | |
+ | Point <math>E</math> is selected on side <math>AB</math> of <math>\triangle{ABC}</math> in such a way that <math>AE: EB = 1: 3</math> and point <math>D</math> is selected on side <math>BC</math> | ||
+ | such that <math>CD: DB = 1: 2</math>. The point of intersection of <math>AD</math> and <math>CE</math> is <math>F</math>. Then <math>\frac {EF}{FC} + \frac {AF}{FD}</math> is: | ||
+ | |||
+ | <math>\textbf{(A)}\ \frac {4}{5} \qquad | ||
+ | \textbf{(B) }\ \frac {5}{4} \qquad | ||
+ | \textbf{(C) }\ \frac {3}{2} \qquad | ||
+ | \textbf{(D) }\ 2\qquad | ||
+ | \textbf{(E) }\ \frac{5}{2} </math> | ||
+ | |||
+ | == Solution == | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | import geometry; | ||
+ | |||
+ | point A = (0,0); | ||
+ | point B = (16,0); | ||
+ | point C = (3, 10); | ||
+ | point D, E, F; | ||
+ | real d; | ||
+ | |||
+ | // Triangle ABC | ||
+ | draw(A--B--C--A); | ||
+ | dot(A); | ||
+ | label("A", A, SW); | ||
+ | dot(B); | ||
+ | label("B", B, SE); | ||
+ | dot(C); | ||
+ | label("C", C, NW); | ||
+ | |||
+ | // Segments AD and CE | ||
+ | D = 2/3*C+1/3*B; | ||
+ | dot(D); | ||
+ | label("D", D, NE); | ||
+ | draw(A--D); | ||
+ | E = midpoint(A--midpoint(A--B)); | ||
+ | dot(E); | ||
+ | label("E", E, S); | ||
+ | draw(C--E); | ||
+ | |||
+ | // Point F | ||
+ | pair[] f=intersectionpoints((A--D), (C--E)); | ||
+ | F=f[0]; | ||
+ | dot(F); | ||
+ | label("F", F, SE); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | We use [[mass points]] for this problem. Let <math>\text{m} A</math> denote the mass of point <math>A</math>. | ||
Rewrite the expression we are finding as | Rewrite the expression we are finding as | ||
<cmath>\frac{EF}{FC} + \frac{AF}{FD} = \frac{FE}{FC} + \frac{FA}{FD} = \frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A}</cmath> | <cmath>\frac{EF}{FC} + \frac{AF}{FD} = \frac{FE}{FC} + \frac{FA}{FD} = \frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A}</cmath> | ||
Line 8: | Line 57: | ||
<cmath>\frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A} = \frac{2}{4} + \frac{3}{3} = \frac{1}{2} + 1 = \frac{3}{2}</cmath> | <cmath>\frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A} = \frac{2}{4} + \frac{3}{3} = \frac{1}{2} + 1 = \frac{3}{2}</cmath> | ||
− | <math>\ | + | This answer corresponds to <math>\fbox{\textbf{(C)}}</math>. |
~JustinLee2017 | ~JustinLee2017 | ||
+ | |||
+ | == See Also == | ||
+ | {{AHSME 40p box|year=1965|num-b=36|num-a=38}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 09:38, 20 July 2024
Problem
Point is selected on side of in such a way that and point is selected on side such that . The point of intersection of and is . Then is:
Solution
We use mass points for this problem. Let denote the mass of point . Rewrite the expression we are finding as Now, let . We then have , so , and We can let . We have From here, substitute the respective values to get
This answer corresponds to .
~JustinLee2017
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 36 |
Followed by Problem 38 | |
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.