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Difference between revisions of "1971 AHSME Problems/Problem 26"

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== Solution 1 ==
 
== Solution 1 ==
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We will use mass points to solve this problem. <math>AC</math> is in the ratio <math>1:2,</math> so we will assign a mass of <math>2</math> to point <math>A,</math> a mass of <math>1</math> to point <math>C,</math> and a mass of <math>3</math> to point <math>F.</math>
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We will use [[mass points]] to solve this problem. <math>AC</math> is in the ratio <math>1:2,</math> so we will assign a mass of <math>2</math> to point <math>A,</math> a mass of <math>1</math> to point <math>C,</math> and a mass of <math>3</math> to point <math>F.</math>
  
 
We also know that <math>G</math> is the midpoint of <math>BF,</math> so <math>BG:GF=1:1.</math> <math>F</math> has a mass of <math>3,</math> so <math>B</math> also has a mass of <math>3.</math>
 
We also know that <math>G</math> is the midpoint of <math>BF,</math> so <math>BG:GF=1:1.</math> <math>F</math> has a mass of <math>3,</math> so <math>B</math> also has a mass of <math>3.</math>

Latest revision as of 19:26, 6 August 2024

Problem

[asy] size(2.5inch); pair A, B, C, E, F, G; A = (0,3); B = (-1,0); C = (3,0); E = (0,0); F = (1,2); G = intersectionpoint(B--F,A--E); draw(A--B--C--cycle); draw(A--E); draw(B--F); label("$A$",A,N); label("$B$",B,W); label("$C$",C,dir(0)); label("$E$",E,S); label("$F$",F,NE); label("$G$",G,SE); //Credit to chezbgone2 for the diagram[/asy]

In $\triangle ABC$, point $F$ divides side $AC$ in the ratio $1:2$. Let $E$ be the point of intersection of side $BC$ and $AG$ where $G$ is the midpoint of $BF$. The point $E$ divides side $BC$ in the ratio

$\textbf{(A) }1:4\qquad \textbf{(B) }1:3\qquad \textbf{(C) }2:5\qquad \textbf{(D) }4:11\qquad \textbf{(E) }3:8$

Solution 1

We will use mass points to solve this problem. $AC$ is in the ratio $1:2,$ so we will assign a mass of $2$ to point $A,$ a mass of $1$ to point $C,$ and a mass of $3$ to point $F.$

We also know that $G$ is the midpoint of $BF,$ so $BG:GF=1:1.$ $F$ has a mass of $3,$ so $B$ also has a mass of $3.$

In line $BC,$ $B$ has a mass of $3$ and $C$ has a mass of $1.$ Therefore, $BE:EC = 1:3.$

The answer is $\boxed{\textbf{(B) }1:3}.$

-edited by coolmath34

Solution 2

By Menelaus' Theorem on $\triangle AGF$ and line $\overleftrightarrow{BC}$, we know that $\tfrac{AE}{GE} \cdot \tfrac{GB}{FB} \cdot \tfrac{FC}{AC}=1$. Because $G$ is the midpoint of $\overline{FB}$, we know that $\tfrac{GB}{FB}=\tfrac12$. Furthermore, from the problem, we know $\tfrac{FC}{AC}=\tfrac23$, so, by susbtitution, our first equation becomes $\tfrac{AE}{GE} \cdot \tfrac12 \cdot \tfrac23 = 1$, so $\tfrac{AE}{GE}=3$, and therefore $\tfrac{AE}{AG}=\tfrac32$.

Using Menelaus again on $\triangle BEG$ and line $\overleftrightarrow{AC}$ reveals that $\tfrac{BC}{EC} \cdot \tfrac{EA}{GA} \cdot \tfrac{GF}{BF}=1$. We know that $\tfrac{EA}{GA}=\tfrac{AE}{AG}=\tfrac32$ and $\tfrac{GF}{BF}=\tfrac12$. Thus, our equation becomes $\tfrac{BC}{EC} \cdot \tfrac32 \cdot \tfrac12=1$, and so $\tfrac{BC}{EC}=\tfrac43$. Because $\tfrac{BC}{EC}=\tfrac{BE+EC}{EC}=1+\tfrac{BE}{EC}$, we see that $\tfrac{BE}{EC}=\tfrac13$, so our answer is $\boxed{\textbf{(B) }1:3}$.

See Also

1971 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
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
All AHSME Problems and Solutions

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