Difference between revisions of "1965 AHSME Problems/Problem 25"
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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: | 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{(B) }\ \text{all sides equal} \\ | ||
\textbf{(C) }\ \text{two pairs of equal sides} \qquad | \textbf{(C) }\ \text{two pairs of equal sides} \qquad | ||
\textbf{(D) }\ \text{one pair of equal sides} \\ | \textbf{(D) }\ \text{one pair of equal sides} \\ | ||
− | \textbf{(E) }\ \text{one pair of equal angles} </math> | + | \textbf{(E) }\ \text{one pair of equal angles} </math> |
== Solution == | == 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 <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 19:52, 18 July 2024
Problem
Let be a quadrilateral with
extended to
so that
.
Lines
and
are drawn to form
. For this angle to be a right angle it is necessary that quadrilateral
have:
Solution
Because is right, the midpoint of its hypoteneuse (namely,
) is its orthocenter. Thus,
, and so two side lengths of quadrilateral
are equal. The placement of
is irrelevant. Thus, our answer is
.
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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