Difference between revisions of "2005 AIME II Problems/Problem 14"
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<math>DK = CK - 6 = 14*15/27 - 6 = 16/9</math>. Thus, <math>\frac{BE}{KE} = \frac{13*54}{14*16}</math>. | <math>DK = CK - 6 = 14*15/27 - 6 = 16/9</math>. Thus, <math>\frac{BE}{KE} = \frac{13*54}{14*16}</math>. | ||
− | Additionally, <math>BE + KE = 9</math>. Solving gives that <math> q = 463.</math> | + | Additionally, <math>BE + KE = 13*15/27 = 65/9</math>. Solving gives that <math> q = 463.</math> |
Alternate: | Alternate: | ||
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</asy></center> | </asy></center> | ||
− | Now we start by drawing altitudes from D and E onto AC, labeling them as F and G, and labelling <math>\angle DAG = \alpha</math>. Now we know that <math>\overline{ | + | Now we start by drawing altitudes from D and E onto AC, labeling them as F and G, and labelling <math>\angle DAG = \alpha</math>. Now we know that <math>\overline{DF} = \frac{24}{5}</math> and <math>\overline{FC} = \frac{18}{5}</math>. Therefore, <math>\overline{AF} = \frac{52}{5}</math>, so <math>\tan{(\alpha)} = \frac{6}{13}</math>. Our goal now is to use tangent <math>\angle EAG</math> in triangle <math>AEG</math>. We set <math>\overline{BE}</math> to <math>x</math>, so <math>\overline{ED} = 9 - x</math> and <math>\overline{EC} = 15 - x</math>, so <math>\overline{EG} = \frac{4}{5}(15-x)</math> and <math>\overline{GC} = \frac{3}{5}(15-x)</math> so <math>\overline{AG} = \frac{3x+25}{5}</math>. Now we just need tangent of <math>\angle EAG</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> | + | 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} \cdot \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 8( | + | == Solution 8 (Isogonal Conjugates) == |
Let <math>ED = x</math>, such that <math>BE = 9-x</math>. Since <math>\overline{AE}</math> and <math>\overline{AD}</math> are isogonal, we get <math>\frac{9-x}{6+x} \cdot \frac{9}{6} = \frac{13^2}{14^2} \Rightarrow 588(9 - x) = 338(6 + x)</math>, and we can solve to get <math>x = \frac{1632}{463}</math>(and <math>BE = \frac{2535}{463}</math>). Hence, our answer is <math>\boxed{463}</math>. - Spacesam | Let <math>ED = x</math>, such that <math>BE = 9-x</math>. Since <math>\overline{AE}</math> and <math>\overline{AD}</math> are isogonal, we get <math>\frac{9-x}{6+x} \cdot \frac{9}{6} = \frac{13^2}{14^2} \Rightarrow 588(9 - x) = 338(6 + x)</math>, and we can solve to get <math>x = \frac{1632}{463}</math>(and <math>BE = \frac{2535}{463}</math>). Hence, our answer is <math>\boxed{463}</math>. - Spacesam | ||
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After getting AJ and BJ, we can get that <math>AB=AJ+BJ=\frac{463x}{28}=13</math>, which means that <math>x=\frac{364}{463}</math>. According to the similarity, <math>\frac{CM}{BC}=\frac{JE}{BE};BE=\frac{2535}{463}</math> which <math>\boxed{463}</math> is our answer | After getting AJ and BJ, we can get that <math>AB=AJ+BJ=\frac{463x}{28}=13</math>, which means that <math>x=\frac{364}{463}</math>. According to the similarity, <math>\frac{CM}{BC}=\frac{JE}{BE};BE=\frac{2535}{463}</math> which <math>\boxed{463}</math> is our answer | ||
~bluesoul | ~bluesoul | ||
+ | |||
+ | == Solution 11 (Ultimate Stewarts Bash) == | ||
+ | First, apply Stewart's theorem to triangle <math>ABC</math> with cevian <math>AD</math>, from which we receive <math>AD=\frac{4\sqrt{41}}{\sqrt{5}}</math>. Then, set <math>AE=y</math> and <math>BE=x.</math> Hence, <math>DE=9-x</math>. Applying Stewart's on triangle <math>ACE</math>, with cevian <math>AD</math>, we receive that <math>y^2-x^2=169-\frac{66}{5}x</math>. By also applying the sine ratio formula on triangles <math>ACD</math> and <math>AEB</math>, since these triangles share the same height, we get that <math>\frac{6}{x}=\frac{\frac{1}{2} \cdot 14 \cdot \frac{4\sqrt{41}}{\sqrt{5}}}{\frac{1}{2} \cdot 13 \cdot y}</math>. Here, we can simplify to receive that <math>\frac{32144}{7605}x^2=y^2</math>. We plug this into our earlier equation, and get that <math>\frac{24539}{7605}x^2=169-\frac{66}{5}x \implies 24539x^2+66 \cdot 39^2x-169 - 169 \cdot 39^2 \cdot 5=0.</math> We then apply the quadratic formula (which may seem computationally intensive, but factoring kills it), and get <math>\frac{-66 \cdot 39^2 + \sqrt{66^2 \cdot 39^4+4(24539)(169 \cdot39^2 \cdot 5)}}{2 \cdot 24539}</math>. (Note we only take the plus symbol instead of <math>\pm</math> since <math>x > 0</math>.) After factoring heavily, we get the answer to be <math>\frac{-33 \cdot 39^2+39 \cdot 13 \cdot 364}{24539} = \frac{2535}{463}</math>, and the answer is <math>\boxed{463}.</math> | ||
+ | |||
+ | ~SirAppel | ||
== See also == | == See also == |
Latest revision as of 16:38, 21 August 2024
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 8 (Isogonal Conjugates)
- 9 Solution 9 (Short and no IQ Required Altogether-Bash)
- 10 Solution 10
- 11 Solution 11 (Ultimate Stewarts Bash)
- 12 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 8 (Isogonal Conjugates)
Let , such that
. Since
and
are isogonal, we get
, and we can solve to get
(and
). Hence, our answer is
. - Spacesam
Solution 9 (Short and no IQ Required Altogether-Bash)
Diagram borrowed from 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)
Applying Law of Cosines on with respect to
we have
Solving gets
, which implies that
Applying Stewart's Theorem with cevian
we have
Solving gets
.
Applying Law of Sines on to solve for
we have
Solving gets
. Thus
.
Applying Law of Sines on we have
Solving gets
.
Applying Stewart's Theorem with cevian we have
Solving gets
Finally, applying Law of Sines on we have
Solving the easy quadratic equation gets
~ Nafer
Solution 10
Making perpendicular lines from to
, meeting at
; from
to
, meeting at
. According to LOC, we can get that
. So we get that
. Now we can see that
. Now we see that in
, assuming
since
. Now we need to find the tangent of
. Making a perpendicular line from
to
at
. We can see that
. We get that
so
.
After getting AJ and BJ, we can get that
, which means that
. According to the similarity,
which
is our answer
~bluesoul
Solution 11 (Ultimate Stewarts Bash)
First, apply Stewart's theorem to triangle with cevian
, from which we receive
. Then, set
and
Hence,
. Applying Stewart's on triangle
, with cevian
, we receive that
. By also applying the sine ratio formula on triangles
and
, since these triangles share the same height, we get that
. Here, we can simplify to receive that
. We plug this into our earlier equation, and get that
We then apply the quadratic formula (which may seem computationally intensive, but factoring kills it), and get
. (Note we only take the plus symbol instead of
since
.) After factoring heavily, we get the answer to be
, and the answer is
~SirAppel
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.