1965 AHSME Problems/Problem 25: Difference between revisions

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 $\fbox{\textbf{(D) }one pair of equal sides}$ .

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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1965 AHSME Problems/Problem 25: Difference between revisions

Latest revision as of 18:52, 18 July 2024

Problem

Solution

See Also

@@ Line 4: / Line 4: @@
 Lines <math>AC</math> and <math>CE</math> are drawn to form <math>\angle{ACE}</math>. For this angle to be a right angle it is necessary that quadrilateral <math>ABCD</math> have:
-<math>\textbf{(A)}\ \text{all angles equal}
+<math>\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} </math>   \qquad
+\textbf{(E) }\ \text{one pair of equal angles} </math>
 == 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 ==