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Difference between revisions of "1965 AHSME Problems/Problem 37"

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(diagram)
 
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== Solution ==
 
== Solution ==
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<asy>
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import geometry;
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point A = (0,0);
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point B = (16,0);
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point C = (3, 10);
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point D, E, F;
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real d;
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// Triangle ABC
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draw(A--B--C--A);
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dot(A);
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label("A", A, SW);
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dot(B);
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label("B", B, SE);
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dot(C);
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label("C", C, NW);
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// Segments AD and CE
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D = 2/3*C+1/3*B;
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dot(D);
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label("D", D, NE);
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draw(A--D);
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E = midpoint(A--midpoint(A--B));
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dot(E);
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label("E", E, S);
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draw(C--E);
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// Point F
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pair[] f=intersectionpoints((A--D), (C--E));
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F=f[0];
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dot(F);
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label("F", F, SE);
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</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>.

Latest revision as of 09:38, 20 July 2024

Problem

Point $E$ is selected on side $AB$ of $\triangle{ABC}$ in such a way that $AE: EB = 1: 3$ and point $D$ is selected on side $BC$ such that $CD: DB = 1: 2$. The point of intersection of $AD$ and $CE$ is $F$. Then $\frac {EF}{FC} + \frac {AF}{FD}$ is:

$\textbf{(A)}\ \frac {4}{5} \qquad \textbf{(B) }\ \frac {5}{4} \qquad \textbf{(C) }\ \frac {3}{2} \qquad \textbf{(D) }\ 2\qquad \textbf{(E) }\ \frac{5}{2}$

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 $\text{m} A$ denote the mass of point $A$. Rewrite the expression we are finding as \[\frac{EF}{FC} + \frac{AF}{FD} = \frac{FE}{FC} + \frac{FA}{FD} = \frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A}\] Now, let $\text{m} C = 2$. We then have $2 \cdot 1 = \text{m} B \cdot 2$, so $\text{m} B = 1$, and $\text{m} D = 2+1 = 3$ We can let $\text{m} A = 3$. We have $\text{m} E = \text{m} A + \text{m} B = 3+1 = 4$ From here, substitute the respective values to get

\[\frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A} = \frac{2}{4} + \frac{3}{3} = \frac{1}{2} + 1 = \frac{3}{2}\] This answer corresponds to $\fbox{\textbf{(C)}}$.

~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

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