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Difference between revisions of "1965 AHSME Problems/Problem 25"

m (fixed typo)
(→‎Solution: with diagram)
 
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== Solution ==
 
== Solution ==
−
<math>\boxed{\textbf{(D)}}</math>
 
  
 +
<asy>
 +
 +
draw((0,0)--(16,0));
 +
dot((0,0));
 +
label("A", (-1,-1));
 +
dot((8,0));
 +
label("B",(8,-1));
 +
dot((16,0));
 +
label("E",(17,-1));
 +
 +
draw((0,0)--(8*sqrt(3),4)--(16,0));
 +
draw((8,0)--(8*sqrt(3),4));
 +
dot((8*sqrt(3),4));
 +
label("C", (8*sqrt(3)+0.75,5));
 +
 +
draw((0,0)--(2,6)--(8*sqrt(3),4));
 +
dot((2,6));
 +
label("D", (1,7));
 +
 +
markscalefactor=0.1;
 +
draw(rightanglemark((0,0), (8*sqrt(3),4), (16,0)));
 +
 +
</asy>
 +
 +
Because <math>\triangle ACE</math> is right, the midpoint of its hypoteneuse (namely, <math>B</math>) is its [[orthocenter]]. Thus, <math>AB=BC</math>, and so two side lengths of quadrilateral <math>ABCD</math> are equal. The placement of <math>D</math> is irrelevant. Thus, our answer is <math>\fbox{\textbf{(D) }one pair of equal sides}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 18:52, 18 July 2024

Problem

Let $ABCD$ be a quadrilateral with $AB$ extended to $E$ so that $\overline{AB} = \overline{BE}$. Lines $AC$ and $CE$ are drawn to form $\angle{ACE}$. For this angle to be a right angle it is necessary that quadrilateral $ABCD$ have:

$\textbf{(A)}\ \text{all angles equal} \qquad \textbf{(B) }\ \text{all sides equal} \\ \textbf{(C) }\ \text{two pairs of equal sides} \qquad \textbf{(D) }\ \text{one pair of equal sides} \\ \textbf{(E) }\ \text{one pair of equal angles}$

Solution

[asy] draw((0,0)--(16,0)); dot((0,0)); label("A", (-1,-1)); dot((8,0)); label("B",(8,-1)); dot((16,0)); label("E",(17,-1)); draw((0,0)--(8*sqrt(3),4)--(16,0)); draw((8,0)--(8*sqrt(3),4)); dot((8*sqrt(3),4)); label("C", (8*sqrt(3)+0.75,5)); draw((0,0)--(2,6)--(8*sqrt(3),4)); dot((2,6)); label("D", (1,7)); markscalefactor=0.1; draw(rightanglemark((0,0), (8*sqrt(3),4), (16,0))); [/asy]

Because $\triangle ACE$ is right, the midpoint of its hypoteneuse (namely, $B$) is its orthocenter. Thus, $AB=BC$, and so two side lengths of quadrilateral $ABCD$ are equal. The placement of $D$ is irrelevant. Thus, our answer is $\fbox{\textbf{(D) }one pair of equal sides}$.

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
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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