Difference between revisions of "1991 AHSME Problems/Problem 19"
m (→Solution 2) |
Jerry122805 (talk | contribs) (→Solution 2) |
||
Line 28: | Line 28: | ||
Extend lines <math>AD</math> and <math>CE</math> to meet at a new point <math>F</math>. Now, we see that <math>FAC\sim FDE \sim ACB</math>. Using this relationship, we can see that <math>AF=\frac{15}4</math>, (so <math>FD=\frac{63}4</math>), and the ratio of similarity between <math>FDE</math> and <math>FAC</math> is <math>\frac{63}{15}</math>. This ratio gives us that <math>\frac{63}5</math>. By the Pythagorean Theorem, <math>DB=13</math>. Thus, <math>\frac{DE}{DB}=\frac{63}{65}</math>, and the answer is <math>63+65=\boxed{\textbf{(B) } 128}</math>. | Extend lines <math>AD</math> and <math>CE</math> to meet at a new point <math>F</math>. Now, we see that <math>FAC\sim FDE \sim ACB</math>. Using this relationship, we can see that <math>AF=\frac{15}4</math>, (so <math>FD=\frac{63}4</math>), and the ratio of similarity between <math>FDE</math> and <math>FAC</math> is <math>\frac{63}{15}</math>. This ratio gives us that <math>\frac{63}5</math>. By the Pythagorean Theorem, <math>DB=13</math>. Thus, <math>\frac{DE}{DB}=\frac{63}{65}</math>, and the answer is <math>63+65=\boxed{\textbf{(B) } 128}</math>. | ||
+ | |||
+ | |||
+ | == Solution 3 (Trig)== | ||
+ | |||
+ | We have <math>\angle ABC = \arcsin(\displaystyle\frac{3}{5})</math> and <math>\angle DBA=\arcsin(\displaystyle\frac{12}{13}).</math> | ||
== See also == | == See also == |
Revision as of 16:24, 21 June 2020
Problem
Triangle has a right angle at
and
. Triangle
has a right angle at
and
. Points
and
are on opposite sides of
. The line through
parallel to
meets
extended at
. If
where
and
are relatively prime positive integers, then
Solution 1
Solution by e_power_pi_times_i
Let be the point such that
and
are parallel to
and
, respectively, and let
and
. Then,
. So,
. Simplifying
, and
. Therefore
, and
. Checking,
is the answer, so
. The answer is
.
Solution 2
Solution by Arjun Vikram
Extend lines and
to meet at a new point
. Now, we see that
. Using this relationship, we can see that
, (so
), and the ratio of similarity between
and
is
. This ratio gives us that
. By the Pythagorean Theorem,
. Thus,
, and the answer is
.
Solution 3 (Trig)
We have and
See also
1991 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.