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

(Created page with "==Solution== 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 <cmath>\frac{E...")
 
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==Solution==
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== Problem ==
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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>
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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:
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<math>\textbf{(A)}\ \frac {4}{5} \qquad
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\textbf{(B) }\ \frac {5}{4} \qquad
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\textbf{(C) }\ \frac {3}{2} \qquad
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\textbf{(D) }\ 2\qquad
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\textbf{(E) }\ \frac{5}{2} </math> 
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== Solution ==
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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>.
 
Rewrite the expression we are finding as  
 
Rewrite the expression we are finding as  

Revision as of 13:54, 16 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

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

\[\frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A} = \frac{2}{4} + \frac{3}{3} = \frac{1}{2} + 1 = \frac{3}{2}\] $\boxed{C}$

~JustinLee2017

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