Difference between revisions of "2005 AIME II Problems/Problem 14"
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We find this using <math>\tan{(EAG)} = \tan{(A - \alpha)} = \frac{\tan{A} - \tan{\alpha}}{1 + \tan{A}\tan{\alpha}}</math>, which is <math>\frac{\frac{12}{5} - \frac{6}{13}}{1 + \frac{12}{5}\frac{6}{13}}</math> or <math>\frac{126}{137}</math>. Now we solve the equation <math>\tan{\angle EAG} = \frac{126}{137} = \frac{\frac{60-4x}{5}}{\frac{3x+25}{5}}</math>, so <math>x = \frac{2535}{463}</math> | We find this using <math>\tan{(EAG)} = \tan{(A - \alpha)} = \frac{\tan{A} - \tan{\alpha}}{1 + \tan{A}\tan{\alpha}}</math>, which is <math>\frac{\frac{12}{5} - \frac{6}{13}}{1 + \frac{12}{5}\frac{6}{13}}</math> or <math>\frac{126}{137}</math>. Now we solve the equation <math>\tan{\angle EAG} = \frac{126}{137} = \frac{\frac{60-4x}{5}}{\frac{3x+25}{5}}</math>, so <math>x = \frac{2535}{463}</math> | ||
+ | |||
+ | == Solution 7 (Super fast solution, 2 billion IQ)== | ||
+ | |||
+ | Let <math><BAE = <CAB = \theta, <EAD = \alpha, BE = x \rightarrow ED = 9-x</math> | ||
+ | |||
+ | Using Ratio Lemma 2 times, we obtain the following equations: | ||
+ | <math> | ||
+ | 1. \tfrac{BD}{DC} = \tfrac{AB}{AC} \cdot \tfrac{\sin (\theta + \alpha)}{\sin (\theta) } | ||
+ | 2. \tfrac{BE}{EC} = \tfrac{AB}{AC} \cdot \tfrac{\sin (\theta)}{\sin (\theta + \alpha)}</math> | ||
+ | |||
+ | Multiplying the two equations, we have <math> \left (\frac{AB}{AC} \right)^2 = \frac{BD \cdot BE}{DC \cdot EC}</math>. Plugging in the values we know, we have <math>\frac{169}{196} = \frac{3x}{30-2x}</math>. Solving for <math>x</math>, we get <math>x = \tfrac{2535}{\boxed{463}}</math> | ||
== See also == | == See also == |
Revision as of 20:16, 31 July 2019
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Similar Triangles)
- 4 Solution 3 (LoC and LoS bash)
- 5 Solution 4 (Ratio Lemma and Angle Bisector Theorem)
- 6 Solution 5 (Isogonal lines with respect to A angle bisesector)
- 7 Solution 6 (Tangent subtraction formulas)
- 8 Solution 7 (Super fast solution, 2 billion IQ)
- 9 See also
Problem
In triangle and
Point
is on
with
Point
is on
such that
Given that
where
and
are relatively prime positive integers, find
Solution 1
![[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("$A$",A,N); label("$B$",B,SW); label("$D$",D,SE); label("$E$",E,S); label("$C$",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]](http://latex.artofproblemsolving.com/f/5/3/f53f3016596f4d0c3c6dc23b6e0b5b41fabe85f2.png)
By the Law of Sines and since , we have
Substituting our knowns, we have . The answer is
.
Solution 2 (Similar Triangles)
Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively.
From here, we can use Heron's Formula to find the altitude. The area of the triangle is . We can then use similar triangles with triangle
and triangle
to find
. Consequently, from Pythagorean theorem,
and
. We can also use the Pythagorean theorem on triangle
to determine that
.
Label as
and
as
.
then equals
. Then, we have two similar triangles.
Firstly: . From there, we have
.
Next: . From there, we have
.
Solve the system to get and
. Notice that 463 is prime, so even though we use the Pythagorean theorem on
and
, the denominator won't change. The answer we desire is
.
Solution 3 (LoC and LoS bash)
Let . Note by Law of Sines on
we have
As a result, our goal is to find
and
(we already know
).
Let the foot of the altitude from to
be
. By law of cosines on
we have
It follows that
and
.
Note that by PT on we have that
. By Law of Sines on
(where we square everything to avoid taking the square root) we see
How are we going to find
though?
and
are in the same triangle. Applying Law of Sines on
we see that
,
, and
are all in the same triangle. We know they add up to
. There's a good chance we can exploit this using the identity
.
We have that . Success! We know
and
already. Applying the
addition formula we see
This is the last stretch! Applying Law of Sines a final time on
we see
It follows that the answer is
.
Solution 4 (Ratio Lemma and Angle Bisector Theorem)
Let be the angle bisector of
such that
is on
.
Then , and thus
.
By the Ratio Lemma,
and
.
This implies that .
Thus, .
. Thus,
.
Additionally, . Solving gives that
Alternate:
By the ratio lemma,
Combining these, we get
Thus,
Solution 5 (Isogonal lines with respect to A angle bisesector)
Since and
are isogonal with respect to the
angle bisector, we have
To prove this, let
and
Then, by the Ratio Lemma, we have
and multiplying these together proves the formula for isogonal lines. Hence, we have
so our desired answer is
Solution 6 (Tangent subtraction formulas)
Note: We first recall some helpful tips regarding 13, 14, 15 triangles. Drawing an altitude H from B to AC results in AHB being a 5-12-13 right triangle and CHB being a 3-4-5 (9-12-15) right triangle.
![[asy] import olympiad; import cse5; import geometry; size(300); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("$A$",A,N); label("$B$",B,SW); label("$D$",D,SE); label("$E$",E,S); label("$C$",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); pair G = foot(E,C,A); pair F = foot(D,C,A); draw(D--F); draw(E--G); label("$G$",G,N); label("$F$",F,N); [/asy]](http://latex.artofproblemsolving.com/b/b/9/bb909db5493f5a40ee3c8271e1848f221d70d213.png)
Now we start by drawing altitudes from D and E onto AC, labeling them as F and G, and labelling . Now we know that
and
. Therefore,
, so
. Our goal now is to use tangent
in triangle
. We set
to
, so
and
, so
and
so
. Now we just need tangent of
.
We find this using , which is
or
. Now we solve the equation
, so
Solution 7 (Super fast solution, 2 billion IQ)
Let
Using Ratio Lemma 2 times, we obtain the following equations:
Multiplying the two equations, we have . Plugging in the values we know, we have
. Solving for
, we get
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.