Difference between revisions of "2005 AIME II Problems/Problem 14"
(→Solution 9 (Long but no IQ Required Altogether Bash)) |
(→Solution 10) |
||
(8 intermediate revisions by 4 users not shown) | |||
Line 57: | Line 57: | ||
Let the foot of the altitude from <math>A</math> to <math>BC</math> be <math>H</math>. By law of cosines on <math>\triangle ABC</math> we have | Let the foot of the altitude from <math>A</math> to <math>BC</math> be <math>H</math>. By law of cosines on <math>\triangle ABC</math> we have | ||
<cmath>169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}</cmath> | <cmath>169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}</cmath> | ||
− | It follows that <math>AH = \frac{56}{5}</math> and <math>HC = \frac{42}{5} \Rightarrow | + | It follows that <math>AH = \frac{56}{5}</math> and <math>HC = \frac{42}{5} \Rightarrow HD = \frac{12}{5}</math>. |
Note that by PT on <math>\triangle AHD</math> we have that <math>AD^2 = \left(\frac{56}{5}\right)^2 + \left(\frac{12}{5}\right)^2 = \frac{656}{5}</math>. By Law of Sines on <math>\triangle ADC</math> (where we square everything to avoid taking the square root) we see | Note that by PT on <math>\triangle AHD</math> we have that <math>AD^2 = \left(\frac{56}{5}\right)^2 + \left(\frac{12}{5}\right)^2 = \frac{656}{5}</math>. By Law of Sines on <math>\triangle ADC</math> (where we square everything to avoid taking the square root) we see | ||
Line 139: | Line 139: | ||
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 8(literally 2 minute solution) == | == Solution 8(literally 2 minute solution) == | ||
Line 156: | Line 145: | ||
− | == Solution 9 ( | + | == Solution 9 (Short and no IQ Required Altogether-Bash) == |
Diagram borrowed from Solution 1. | Diagram borrowed from Solution 1. | ||
Line 200: | Line 189: | ||
<cmath>(BC)(AE)^2+(BC)(BE)(CE)=(CE)(AB)^2+(BE)(AC)^2</cmath> | <cmath>(BC)(AE)^2+(BC)(BE)(CE)=(CE)(AB)^2+(BE)(AC)^2</cmath> | ||
<cmath>(BC)(AE)^2+(BC)(BE)(BC-BE)=(BC-BE)(AB)^2+(BE)(AC)^2</cmath> | <cmath>(BC)(AE)^2+(BC)(BE)(BC-BE)=(BC-BE)(AB)^2+(BE)(AC)^2</cmath> | ||
− | Solving gets <math>AE=\sqrt{\frac{15BE^2-198BE+2535}{15}</math> | + | Solving gets <math>AE=\sqrt{\frac{15BE^2-198BE+2535}{15}}</math> |
− | Finally, applying [[Law of Sines]] on <math>\ | + | Finally, applying [[Law of Sines]] on <math>\bigtriangleup BAE</math> we have |
<cmath>\frac{AE}{\sin B}=\frac{BE}{\sin BAE}</cmath> | <cmath>\frac{AE}{\sin B}=\frac{BE}{\sin BAE}</cmath> | ||
− | <cmath>\frac{\sqrt{\frac{15BE^2-198BE+2535}{15}}{\frac{56}{65}}=\frac{BE}{\frac{6\sqrt{205}}{205}}</cmath> | + | <cmath>\frac{\sqrt{\frac{15BE^2-198BE+2535}{15}}}{\frac{56}{65}}=\frac{BE}{\frac{6\sqrt{205}}{205}}</cmath> |
<cmath>7605BE^2-32342BE+2535=0</cmath> | <cmath>7605BE^2-32342BE+2535=0</cmath> | ||
− | Solving | + | Solving the easy quadratic equation gets <math>BE=\frac{1632}{463}\Longrightarrow q=\boxed{463}</math> |
+ | |||
+ | ~ Nafer | ||
+ | == Solution 10 == | ||
+ | Making perpendicular lines from <math>D</math> to <math>AC</math>, meeting at <math>N</math>; from <math>E</math> to <math>AB</math>, meeting at <math>J</math>. According to LOC, we can get that <math>cos\angle{C}=\frac{3}{5}</math>. So we get that <math>CN=\frac{18}{5};DN=\frac{24}{5};AN=AC-CN=\frac{52}{5}</math>. Now we can see that <math>tan\angle{DAN}=\frac{DN}{AN}=\frac{6}{13}</math>. Now we see that in <math>\triangle{EJA}</math>, assuming <math>EJ=6x;AJ=13x</math> since <math>tan\angle{JAE}=tan\angle{DAN}</math>. Now we need to find the tangent of <math>\angle{B}</math>. Making a perpendicular line from <math>C</math> to <math>AB</math> at <math>M</math>. We can see that <math>CM=\frac{168}{13};AM=\frac{70}{13};BM=AB-AM=13-\frac{70}{13}=\frac{99}{13}</math>. We get that <math>tan\angle{B}=\frac{56}{33}</math> so <math>BJ=\frac{33}{56}*6x=\frac{99x}{28}</math>. | ||
+ | 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 | ||
== See also == | == See also == |
Latest revision as of 02:46, 31 October 2021
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(literally 2 minute solution)
- 9 Solution 9 (Short and no IQ Required Altogether-Bash)
- 10 Solution 10
- 11 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
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.
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(literally 2 minute solution)
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.
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
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.