Difference between revisions of "1971 AHSME Problems/Problem 26"
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In <math>\triangle ABC</math>, point <math>F</math> divides side <math>AC</math> in the ratio <math>1:2</math>. Let <math>E</math> be the point of intersection of side <math>BC</math> and <math>AG</math> where <math>G</math> is the | In <math>\triangle ABC</math>, point <math>F</math> divides side <math>AC</math> in the ratio <math>1:2</math>. Let <math>E</math> be the point of intersection of side <math>BC</math> and <math>AG</math> where <math>G</math> is the | ||
− | + | midpoint of <math>BF</math>. The point <math>E</math> divides side <math>BC</math> in the ratio | |
<math>\textbf{(A) }1:4\qquad | <math>\textbf{(A) }1:4\qquad | ||
Line 30: | Line 30: | ||
\textbf{(E) }3:8 </math> | \textbf{(E) }3:8 </math> | ||
− | == Solution == | + | == Solution 1 == |
− | 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 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> | ||
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In line <math>BC,</math> <math>B</math> has a mass of <math>3</math> and <math>C</math> has a mass of <math>1.</math> Therefore, <math>BE:EC = 1:3.</math> | In line <math>BC,</math> <math>B</math> has a mass of <math>3</math> and <math>C</math> has a mass of <math>1.</math> Therefore, <math>BE:EC = 1:3.</math> | ||
− | The answer is <math>\textbf{(B)}.</math> | + | The answer is <math>\boxed{\textbf{(B) }1:3}.</math> |
-edited by coolmath34 | -edited by coolmath34 | ||
+ | |||
+ | == Solution 2 == | ||
+ | By [[Menelaus' Theorem]] on <math>\triangle AGF</math> and line <math>\overleftrightarrow{BC}</math>, we know that <math>\tfrac{AE}{GE} \cdot \tfrac{GB}{FB} \cdot \tfrac{FC}{AC}=1</math>. Because <math>G</math> is the midpoint of <math>\overline{FB}</math>, we know that <math>\tfrac{GB}{FB}=\tfrac12</math>. Furthermore, from the problem, we know <math>\tfrac{FC}{AC}=\tfrac23</math>, so, by susbtitution, our first equation becomes <math>\tfrac{AE}{GE} \cdot \tfrac12 \cdot \tfrac23 = 1</math>, so <math>\tfrac{AE}{GE}=3</math>, and therefore <math>\tfrac{AE}{AG}=\tfrac32</math>. | ||
+ | |||
+ | Using Menelaus again on <math>\triangle BEG</math> and line <math>\overleftrightarrow{AC}</math> reveals that <math>\tfrac{BC}{EC} \cdot \tfrac{EA}{GA} \cdot \tfrac{GF}{BF}=1</math>. We know that <math>\tfrac{EA}{GA}=\tfrac{AE}{AG}=\tfrac32</math> and <math>\tfrac{GF}{BF}=\tfrac12</math>. Thus, our equation becomes <math>\tfrac{BC}{EC} \cdot \tfrac32 \cdot \tfrac12=1</math>, and so <math>\tfrac{BC}{EC}=\tfrac43</math>. Because <math>\tfrac{BC}{EC}=\tfrac{BE+EC}{EC}=1+\tfrac{BE}{EC}</math>, we see that <math>\tfrac{BE}{EC}=\tfrac13</math>, so our answer is <math>\boxed{\textbf{(B) }1:3}</math>. | ||
+ | |||
+ | == See Also == | ||
+ | {{AHSME 35p box|year=1971|num-b=25|num-a=27}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:26, 6 August 2024
Contents
Problem
In , point
divides side
in the ratio
. Let
be the point of intersection of side
and
where
is the
midpoint of
. The point
divides side
in the ratio
Solution 1
We will use mass points to solve this problem. is in the ratio
so we will assign a mass of
to point
a mass of
to point
and a mass of
to point
We also know that is the midpoint of
so
has a mass of
so
also has a mass of
In line
has a mass of
and
has a mass of
Therefore,
The answer is
-edited by coolmath34
Solution 2
By Menelaus' Theorem on and line
, we know that
. Because
is the midpoint of
, we know that
. Furthermore, from the problem, we know
, so, by susbtitution, our first equation becomes
, so
, and therefore
.
Using Menelaus again on and line
reveals that
. We know that
and
. Thus, our equation becomes
, and so
. Because
, we see that
, so our answer is
.
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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