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 18:52, 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
![[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]](http://latex.artofproblemsolving.com/b/0/8/b084d23b093a0968b0c9172c5e5f80a5b95a7705.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
