Difference between revisions of "1965 AHSME Problems/Problem 37"
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== Solution == | == 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>. | We use [[mass points]] for this problem. Let <math>\text{m} A</math> denote the mass of point <math>A</math>. | ||
Line 20: | 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> | ||
− | This answer corresponds to | + | This answer corresponds to <math>\fbox{\textbf{(C)}}</math>. |
~JustinLee2017 | ~JustinLee2017 | ||
== See Also == | == See Also == | ||
− | {{AHSME 40p box|year=1965|num-b= | + | {{AHSME 40p box|year=1965|num-b=36|num-a=38}} |
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 10:38, 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.