Difference between revisions of "1965 AHSME Problems/Problem 37"
m (see also box, small misc changes) |
(diagram) |
||
| (2 intermediate revisions by the same user not shown) | |||
| Line 11: | Line 11: | ||
== 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 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
![[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]](http://latex.artofproblemsolving.com/2/7/1/2714544397c280b137d2054bd7d65d2f867baaa6.png)
We use mass points for this problem. Let 
denote the mass of point 
.
Rewrite the expression we are finding as
![\[\frac{EF}{FC} + \frac{AF}{FD} = \frac{FE}{FC} + \frac{FA}{FD} = \frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A}\]](http://latex.artofproblemsolving.com/f/3/a/f3a717b11449847a3c77699b0f4bd4dad3cefce6.png)
Now, let 
. We then have 
, so 
, and 
We can let 
. We have 
From here, substitute the respective values to get
![\[\frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A} = \frac{2}{4} + \frac{3}{3} = \frac{1}{2} + 1 = \frac{3}{2}\]](http://latex.artofproblemsolving.com/e/0/4/e0476efaade33366dd29f64d3c2c9f4ee7095498.png)
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
