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Difference between revisions of "1951 AHSME Problems/Problem 27"
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1.An equilateral triangle with side lengths <math>x</math>
 
1.An equilateral triangle with side lengths <math>x</math>
  
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2.An isosceles triangle with sidelengths <math>x+3,x+3,x-1</math>
+
2.An isosceles triangle with side lengths <math>x+3,x+3,x-1</math>
  
 
3.A scalene triangle with sides <math>x+7,x-3,x+4</math>
 
3.A scalene triangle with sides <math>x+7,x-3,x+4</math>
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<math>\textbf{(B)}</math>-Since all the triangles are congruent they must always be congruent in any way.
 
<math>\textbf{(B)}</math>-Since all the triangles are congruent they must always be congruent in any way.
  
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<math>\textbf{(C)}</math>-The triangles are all congruent meaning that they al have equal area.
+
<math>\textbf{(C)}</math>-The triangles are all congruent meaning that they always have equal area.
  
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<math>\textbf{(D)}</math>-The top <math>2</math> triangles, bottom-right <math>2</math> triangles, and bottom-left <math>2</math> triangles all make 3 identical congreunt quadrilaterals.
+
<math>\textbf{(D)}</math>-The top <math>2</math> triangles, bottom-right <math>2</math> triangles, and bottom-left <math>2</math> triangles all make 3 identical congruent quadrilaterals.
  
 
2. Gives us 3 pairs of adjacent congruent triangles, which means
 
2. Gives us 3 pairs of adjacent congruent triangles, which means
Line 35: Line 35:
 
<math>\textbf{(E)}</math>-All of the relationships don't work so we must choose the last choice, no relations.
 
<math>\textbf{(E)}</math>-All of the relationships don't work so we must choose the last choice, no relations.
  
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3. Is not needed as we have found that all of the relationships given are not neccesary all the time, so <math>\textbf(E)</math> is the only choice that works.
+
3. Is not needed as we have found that all of the relationships given are not necessary all the time, so <math>\textbf(E)</math> is the only choice that works.
  
 +
So the answer is  <math>\fbox{\textbf{(E)}}</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 15:48, 10 September 2018

Problem

Through a point inside a triangle, three lines are drawn from the vertices to the opposite sides forming six triangular sections. Then:

$\textbf{(A)}\ \text{the triangles are similar in opposite pairs}\qquad\textbf{(B)}\ \text{the triangles are congruent in opposite pairs}$ $\textbf{(C)}\ \text{the triangles are equal in area in opposite pairs}\qquad\textbf{(D)}\ \text{three similar quadrilaterals are formed}$ $\textbf{(E)}\ \text{none of the above relations are true}$

Solution

Say we draw 3 different types of triangles as the following

1.An equilateral triangle with side lengths $x$

2.An isosceles triangle with side lengths $x+3,x+3,x-1$

3.A scalene triangle with sides $x+7,x-3,x+4$

1. Gives us 6 congruent $30-60-90$ triangles, which means

$\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)}$ are all true. therefore 1. does not give any unique criteria.

$\textbf{(A)}$-Since the triangles are congruent with each other, they are similar by a ratio of $1$.

$\textbf{(B)}$-Since all the triangles are congruent they must always be congruent in any way.

$\textbf{(C)}$-The triangles are all congruent meaning that they always have equal area.

$\textbf{(D)}$-The top $2$ triangles, bottom-right $2$ triangles, and bottom-left $2$ triangles all make 3 identical congruent quadrilaterals.

2. Gives us 3 pairs of adjacent congruent triangles, which means

$\textbf{(E)}$ is the only choice that works.

$\textbf{(E)}$-All of the relationships don't work so we must choose the last choice, no relations.

3. Is not needed as we have found that all of the relationships given are not necessary all the time, so $\textbf(E)$ is the only choice that works.

So the answer is $\fbox{\textbf{(E)}}$

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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All AHSME Problems and Solutions

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