Difference between revisions of "1986 AIME Problems/Problem 3"
(→Solution) |
(New solution) |
||
Line 24: | Line 24: | ||
<math>\frac{25}{30} = \frac{\frac{\sin(x+y)}{\cos x \cos y}}{\frac{\sin(x+y)}{\sin x \sin y}} = \frac{\sin x \sin y}{\cos x \cos y} = \tan x \tan y = \frac{5}{6}</math>. Plugging in to the earlier formula, we have <math>\tan(x+y) = \frac{25}{1-\frac{5}{6}} = \frac{25}{\frac{1}{6}} = \boxed{150}</math>. | <math>\frac{25}{30} = \frac{\frac{\sin(x+y)}{\cos x \cos y}}{\frac{\sin(x+y)}{\sin x \sin y}} = \frac{\sin x \sin y}{\cos x \cos y} = \tan x \tan y = \frac{5}{6}</math>. Plugging in to the earlier formula, we have <math>\tan(x+y) = \frac{25}{1-\frac{5}{6}} = \frac{25}{\frac{1}{6}} = \boxed{150}</math>. | ||
+ | |||
+ | == Solution 3 (less trig required, use of quadratic formula) == | ||
+ | |||
+ | Let <math>a=tan(x)</math> and <math>b=tan(y)</math>. This simplifies the equations to: | ||
+ | |||
+ | <cmath>a + b = 25</cmath> | ||
+ | |||
+ | <cmath>\frac{1}{a} + \frac{1}{b} = 30</cmath> | ||
+ | |||
+ | Taking the tangent of a sum formula from Solution 2, we get <math>\frac{25}{1 - ab}</math>. | ||
+ | |||
+ | We can use substitution to solve the system of equations. <math>b = -a + 25</math>, so <math>\frac{1}{a} + \frac{1}{-a + 25} = 30</math>. | ||
+ | |||
+ | Multiplying by <math>-a(a-25)</math>, we get <math>a + (-a + 25) = -30a(a-25)</math>, which is <math>-30a^2 + 750a = 25</math>. Dividing everything by 5 and shifting everything to one side gives <math>6a^2 - 150a + 5 = 0</math>. | ||
+ | |||
+ | Using the quadratic formula gives <math>a = \frac{150 \pm \sqrt {22380}}{12}</math>. Since this looks too hard to simplify, we can solve for <math>b</math> using <math>a + b = 25</math>, which turns out to also be <math>b = \frac{150 \pm \sqrt {22380}}{12}</math>, provided that the sign of the radical in <math>a</math> is opposite the one in <math>b</math>. | ||
+ | |||
+ | WLOG, assume <math>a = \frac{150 + \sqrt{22380}}{12}</math> and <math>b = \frac{150 - \sqrt{22380}}{12}</math>. Multiplying them gives <math>ab = \frac{22500 - 22380}{144}</math> which simplifies to <math>\frac{5}{6}</math>. | ||
+ | |||
+ | THe denominator of <math>\frac{25}{1 - ab}</math> ends up being <math>\frac{1}{6}</math>, so multiplying both numerator and denominator by 6 gives <math>\boxed{150}</math>. | ||
+ | |||
+ | -ThisUsernameIsTaken | ||
== See also == | == See also == |
Revision as of 17:27, 13 March 2021
Contents
Problem
If and
, what is
?
Solution 1
Since is the reciprocal function of
:
Thus,
Using the tangent addition formula:
.
Solution 2
Using the formula for tangent of a sum, . We only need to find
.
We know that . Cross multiplying, we have
.
Similarly, we have .
Dividing:
. Plugging in to the earlier formula, we have
.
Solution 3 (less trig required, use of quadratic formula)
Let and
. This simplifies the equations to:
Taking the tangent of a sum formula from Solution 2, we get .
We can use substitution to solve the system of equations. , so
.
Multiplying by , we get
, which is
. Dividing everything by 5 and shifting everything to one side gives
.
Using the quadratic formula gives . Since this looks too hard to simplify, we can solve for
using
, which turns out to also be
, provided that the sign of the radical in
is opposite the one in
.
WLOG, assume and
. Multiplying them gives
which simplifies to
.
THe denominator of ends up being
, so multiplying both numerator and denominator by 6 gives
.
-ThisUsernameIsTaken
See also
1986 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.