https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Hiker&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T15:05:10ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_14&diff=1229332005 AIME II Problems/Problem 142020-05-25T03:51:05Z<p>Hiker: /* Solution 7 (Super fast solution, 2 billion IQ) */</p>
<hr />
<div>== Problem ==<br />
<br />
In [[triangle]] <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such that <math> \angle BAE\cong \angle CAD. </math> Given that <math> BE=\frac pq </math> where <math> p </math> and <math> q </math> are relatively prime positive integers, find <math> q. </math><br />
<br />
== Solution 1==<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(150);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
</asy></center><br />
<br />
By the [[Law of Sines]] and since <math>\angle BAE = \angle CAD, \angle BAD = \angle CAE</math>, we have<br />
<br />
<cmath>\begin{align*}<br />
\frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\<br />
&= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\<br />
&= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2}<br />
\end{align*}<br />
</cmath><br />
<br />
Substituting our knowns, we have <math>\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}</math>. The answer is <math>q = \boxed{463}</math>.<br />
<br />
== Solution 2 (Similar Triangles)==<br />
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. <br />
<br />
From here, we can use Heron's Formula to find the altitude. The area of the triangle is <math>\sqrt{21*6*7*8} = 84</math>. We can then use similar triangles with triangle <math>AQC</math> and triangle <math>DSC</math> to find <math>DS=\frac{24}{5}</math>. Consequently, from Pythagorean theorem, <math>SC = \frac{18}{5}</math> and <math>AS = 14-SC = \frac{52}{5}</math>. We can also use the Pythagorean theorem on triangle <math>AQB</math> to determine that <math>BQ = \frac{33}{5}</math>.<br />
<br />
Label <math>AR</math> as <math>y</math> and <math>RE</math> as <math>x</math>. <math>RB</math> then equals <math>13-y</math>. Then, we have two similar triangles. <br />
<br />
Firstly: <math>\triangle ARE \sim \triangle ASD</math>. From there, we have <math>\frac{x}{y} = \frac{\frac{24}{5}}{\frac{53}{5}}</math>. <br />
<br />
Next: <math>\triangle BRE \sim \triangle BQA</math>. From there, we have <math>\frac{x}{13-y} = \frac{\frac{56}{5}}{\frac{33}{5}}</math>.<br />
<br />
Solve the system to get <math>x = \frac{2184}{463}</math> and <math>y = \frac{4732}{463}</math>. Notice that 463 is prime, so even though we use the Pythagorean theorem on <math>x</math> and <math>13-y</math>, the denominator won't change. The answer we desire is <math>\boxed{463}</math>.<br />
<br />
==Solution 3 (LoC and LoS bash)==<br />
Let <math>\angle CAD = \angle BAE = \theta</math>. Note by Law of Sines on <math>\triangle BEA</math> we have<br />
<cmath>\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{\angle BEA}}</cmath><br />
As a result, our goal is to find <math>\sin{\angle BEA}</math> and <math>\sin{\theta}</math> (we already know <math>AB</math>). <br />
<br />
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<br />
<cmath>169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}</cmath><br />
It follows that <math>AH = \frac{56}{5}</math> and <math>HC = \frac{42}{5} \Rightarrow AD = \frac{12}{5}</math>. <br />
<br />
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<br />
<cmath>\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}.</cmath><br />
How are we going to find <math>\sin{\angle BEA}</math> though? <math>\angle BEA</math> and <math>\theta</math> are in the same triangle. Applying Law of Sines on <math>\triangle ABC</math> we see that<br />
<cmath>\frac{13}{\frac{4}{5}} = \frac{14}{\sin{\angle B}} \Rightarrow \sin{\angle B} = \frac{56}{65} \Rightarrow \cos{\angle B} = \frac{33}{65}.</cmath><br />
<math>\theta</math>, <math>\angle B</math>, and <math>\angle BEA</math> are all in the same triangle. We know they add up to <math>180^{\circ}</math>. There's a good chance we can exploit this using the identity <math>\sin{p} = \sin{180^{\circ}-p}</math>.<br />
<br />
We have that <math>\sin{(180^{\circ} - (\theta + \angle B))} = \sin{\angle BEA} = \sin{(\theta + \angle B)}</math>. Success! We know <math>\sin{\theta}</math> and <math>\sin{\angle B}</math> already. Applying the <math>\sin</math> addition formula we see<br />
<cmath>\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}=\frac{1}{65 \cdot \sqrt{205}} (198 + 728) = \frac{926}{65 \sqrt{205}}.</cmath><br />
This is the last stretch! Applying Law of Sines a final time on <math>\triangle BEA</math> we see<br />
<cmath>\frac{BE}{\sin{\theta}} = \frac{13}{\sin{BEA}} \Rightarrow \frac{BE}{\frac{6}{\sqrt{205}}} = \frac{13}{\frac{926}{65\sqrt{205}}} \Rightarrow \frac{BE}{6} = \frac{13 \cdot 65}{926} \Rightarrow \frac{13 \cdot 65 \cdot 6}{926} = BE = \frac{2535}{463}.</cmath><br />
It follows that the answer is <math>\boxed{463}</math>.<br />
<br />
==Solution 4 (Ratio Lemma and Angle Bisector Theorem)==<br />
<br />
Let <math>AK</math> be the angle bisector of <math>\angle A</math> such that <math>K</math> is on <math>BC</math>.<br />
<br />
Then <math>\angle KAB = \angle KAC</math>, and thus <math>\angle KAE = \angle KAD</math>.<br />
<br />
By the Ratio Lemma, <br />
<math>\frac{BE}{KE} = \frac{BA}{KA} * \frac{\sin{BAE}}{\sin{KAE}}</math> and <math>\frac{CD}{KD} = \frac{CA}{KA} * \frac{\sin{CAD}}{\sin{KAD}}</math>.<br />
<br />
This implies that <math>\frac{BE}{KE*BA} = \frac{CD}{KD*CA}</math>.<br />
<br />
Thus, <math>\frac{BE}{KE} = \frac{13}{14} * \frac{6}{DK}</math>.<br />
<br />
<math>DK = CK - 6 = 14*15/27 - 6 = 16/9</math>. Thus, <math>\frac{BE}{KE} = \frac{13*54}{14*16}</math>.<br />
<br />
Additionally, <math>BE + KE = 9</math>. Solving gives that <math> q = 463.</math><br />
<br />
Alternate:<br />
By the ratio lemma,<br />
<math>BD/DC = (13/14)*(\sin BAD/\sin DAC)</math><br />
<math>EC/EB = (14/13)*(\sin EAC/\sin BAE)</math><br />
<br />
Combining these, we get<br />
<math>(BD/DC)(14/13) = (EC/EB)(13/14)</math><br />
<math>(3/2)(14/13)(14/13) = (15-x)(x)</math><br />
<br />
<math>x = 2535/463</math><br />
Thus, <math>q = 463</math><br />
<br />
==Solution 5 (Isogonal lines with respect to A angle bisesector)==<br />
Since <math>AE</math> and <math>AD</math> are isogonal with respect to the <math>A</math> angle bisector, we have <cmath>\frac{BE}{EC}\cdot \frac{BD}{DC}=(\frac{AB}{AC})^2.</cmath> To prove this, let <math>\angle BAE=\angle DAC=x</math> and <math>\angle BAD=\angle CAE=y.</math> Then, by the Ratio Lemma, we have <cmath> \frac{BD}{DC}=\frac{AB\sin y}{AC\sin x}</cmath> <cmath> \frac{BE}{EC}=\frac{AB\sin x}{AC\sin y}</cmath> and multiplying these together proves the formula for isogonal lines. Hence, we have <cmath>\frac{BE}{15-BE}\cdot \frac{9}{6}=\frac{169}{196}\implies BE=\frac{2535}{463}</cmath> so our desired answer is <math>\boxed{463}.</math><br />
<br />
<br />
<br />
==Solution 6 (Tangent subtraction formulas)==<br />
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. <br />
<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(300);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
pair G = foot(E,C,A);<br />
pair F = foot(D,C,A);<br />
draw(D--F);<br />
draw(E--G);<br />
label("$G$",G,N);<br />
label("$F$",F,N);<br />
</asy></center><br />
<br />
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{DG} = \frac{24}{5}</math> and <math>\overline{GC} = \frac{18}{5}</math>. Therefore, <math>\overline{AG} = \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>. <br />
<br />
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><br />
<br />
== Solution 8(literally 2 minute solution) ==<br />
<br />
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{1146}{463}</math>). Hence, our answer is <math>\boxed{463}</math>. - Spacesam<br />
<br />
<br />
== Solution 9 (Short and no IQ Required Altogether-Bash) ==<br />
<br />
Diagram borrowed from Solution 1.<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(150);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
</asy></center><br />
<br />
Applying [[Law of Cosines]] on <math>\bigtriangleup ABC</math> with respect to <math>\angle C</math> we have<br />
<cmath>AB^2=AC^2+BC^2-2(AC)(BC)\cos C</cmath><br />
Solving gets <math>\cos C=\frac{3}{5}</math>, which implies that <br />
<cmath>\sin C=\sqrt{1-\cos C}=\frac{4}{5}</cmath><br />
Applying [[Stewart's Theorem]] with [[cevian]] <math>AD</math> we have<br />
<cmath>(BC)(AD)^2+(BC)(BD)(CD)=(CD)(AB)^2+(BD)(AC)^2</cmath><br />
Solving gets <math>AD=\frac{4\sqrt{205}}{5}</math>.<br />
<br />
Applying [[Law of Sines]] on <math>\bigtriangleup ACD</math> to solve for <math>\sin CAD</math> we have<br />
<cmath>\frac{AD}{\sin C}=\frac{CD}{\sin CAD}</cmath><br />
Solving gets <math>\sin CAD=\frac{6\sqrt{205}}{205}</math>. Thus <math>\sin BAE=\sin CAD=\frac{6\sqrt{205}}{205}</math>.<br />
<br />
Applying [[Law of Sines]] on <math>\bigtriangleup ABC</math> we have<br />
<cmath>\frac{AC}{\sin B}=\frac{AB}{\sin C}</cmath><br />
Solving gets <math>\sin B=\frac{56}{65}</math>.<br />
<br />
Applying [[Stewart's Theorem]] with [[cevian]] <math>AE</math> we have<br />
<cmath>(BC)(AE)^2+(BC)(BE)(CE)=(CE)(AB)^2+(BE)(AC)^2</cmath><br />
<cmath>(BC)(AE)^2+(BC)(BE)(BC-BE)=(BC-BE)(AB)^2+(BE)(AC)^2</cmath><br />
Solving gets <math>AE=\sqrt{\frac{15BE^2-198BE+2535}{15}}</math><br />
<br />
Finally, applying [[Law of Sines]] on <math>\bigtriangleup BAE</math> we have<br />
<cmath>\frac{AE}{\sin B}=\frac{BE}{\sin BAE}</cmath><br />
<cmath>\frac{\sqrt{\frac{15BE^2-198BE+2535}{15}}}{\frac{56}{65}}=\frac{BE}{\frac{6\sqrt{205}}{205}}</cmath><br />
<cmath>7605BE^2-32342BE+2535=0</cmath><br />
Solving the easy quadratic equation gets <math>BE=\frac{1632}{463}\Longrightarrow q=\boxed{463}</math><br />
<br />
~ Nafer<br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=II|num-b=13|num-a=15|t=368562}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_11&diff=1207192015 AIME II Problems/Problem 112020-04-08T18:40:20Z<p>Hiker: /* Solution 7 */</p>
<hr />
<div>==Problem==<br />
<br />
The circumcircle of acute <math>\triangle ABC</math> has center <math>O</math>. The line passing through point <math>O</math> perpendicular to <math>\overline{OB}</math> intersects lines <math>AB</math> and <math>BC</math> and <math>P</math> and <math>Q</math>, respectively. Also <math>AB=5</math>, <math>BC=4</math>, <math>BQ=4.5</math>, and <math>BP=\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Diagram==<br />
<asy><br />
unitsize(30);<br />
draw(Circle((0,0),3));<br />
pair A,B,C,O, Q, P, M, N;<br />
A=(2.5, -sqrt(11/4));<br />
B=(-2.5, -sqrt(11/4));<br />
C=(-1.96, 2.28);<br />
Q=(-1.89, 2.81);<br />
P=(1.13, -1.68);<br />
O=origin;<br />
M=foot(O,C,B);<br />
N=foot(O,A,B);<br />
draw(A--B--C--cycle);<br />
label("$A$",A,SE);<br />
label("$B$",B,SW);<br />
label("$C$",C,NW);<br />
label("$Q$",Q,NW);<br />
dot(O);<br />
label("$O$",O,NE);<br />
label("$M$",M,W);<br />
label("$N$",N,S);<br />
label("$P$",P,S);<br />
draw(B--O);<br />
draw(C--Q);<br />
draw(Q--O);<br />
draw(O--C);<br />
draw(O--A);<br />
draw(O--P);<br />
draw(O--M, dashed);<br />
draw(O--N, dashed);<br />
draw(rightanglemark((-2.5, -sqrt(11/4)),(0,0),(-1.89, 2.81),5));<br />
draw(rightanglemark(O,N,B,5));<br />
draw(rightanglemark(B,O,P,5));<br />
draw(rightanglemark(O,M,C,5));<br />
</asy><br />
===Solution 1===<br />
Call the <math>M</math> and <math>N</math> foot of the altitudes from <math>O</math> to <math>BC</math> and <math>AB</math>, respectively. Let <math>OB = r</math> . Notice that <math>\triangle{OMB} \sim \triangle{QOB}</math> because both are right triangles, and <math>\angle{OBQ} \cong \angle{OBM}</math>. By <math>\frac{MB}{BO}=\frac{BO}{BQ}</math>, <math>MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}</math>. However, since <math>O</math> is the circumcenter of triangle <math>ABC</math>, <math>OM</math> is a perpendicular bisector by the definition of a circumcenter. Hence, <math>\frac{r^2}{4.5} = 2 \implies r = 3</math>. Since we know <math>BN=\frac{5}{2}</math> and <math>\triangle BOP \sim \triangle NBO</math>, we have <math>\frac{BP}{3} = \frac{3}{\frac{5}{2}}</math>. Thus, <math>BP = \frac{18}{5}</math>. <math>m + n=\boxed{023}</math>.<br />
<br />
===Solution 2===<br />
Notice that <math>\angle{CBO}=90-A</math>, so <math>\angle{BQO}=A</math>. From this we get that <math>\triangle{BPQ}\sim \triangle{BCA}</math>. So <math>\dfrac{BP}{BC}=\dfrac{BQ}{BA}</math>, plugging in the given values we get <math>\dfrac{BP}{4}=\dfrac{4.5}{5}</math>, so <math>BP=\dfrac{18}{5}</math>, and <math>m+n=\boxed{023}</math>.<br />
<br />
===Solution 3===<br />
Let <math>r=BO</math>. Drawing perpendiculars, <math>BM=MC=2</math> and <math>BN=NA=2.5</math>. From there, <math>OM=\sqrt{r^2-4}</math>. Thus, <math>OQ=\frac{\sqrt{4r^2+9}}{2}</math>. Using <math>\triangle{BOQ}</math>, we get <math>r=3</math>. Now let's find <math>NP</math>. After some calculations with <math>\triangle{BON}</math> ~ <math>\triangle{OPN}</math>, <math>{NP=11/10}</math>. Therefore, <math>BP=\frac{5}{2}+\frac{11}{10}=18/5</math>. <math>18+5=\boxed{023}</math>.<br />
<br />
===Solution 4===<br />
Let <math>\angle{BQO}=\alpha</math>. Extend <math>OB</math> to touch the circumcircle at a point <math>K</math>. Then, note that <math>\angle{KAC}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha</math>. But since <math>BK</math> is a diameter, <math>\angle{KAB}=90^\circ</math>, implying <math>\angle{CAB}=\alpha</math>. It follows that <math>APCQ</math> is a cyclic quadrilateral.<br />
<br />
Let <math>BP=x</math>. By Power of a Point, <cmath>5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.</cmath>The answer is <math>18+5=\boxed{023}</math>.<br />
<br />
===Solution 5===<br />
Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.<br />
<br />
Denote the circumradius of <math>ABC</math> to be <math>R</math>, the circumcircle of <math>ABC</math> to be <math>O</math>, and the shortest distance from <math>Q</math> to circle <math>O</math> to be <math>x</math>. <br />
<br />
Using Power of a Point on <math>Q</math> relative to circle <math>O</math>, we get that <math>x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}</math>. Using Pythagorean Theorem on triangle <math>QOB</math> to get <math>(x + r)^2 + r^2 = \frac{81}{4}</math>. Subtracting the first equation from the second, we get that <math>2r^2 = 18</math> and therefore <math>r = 3</math>. Now, set <math>\cos{ABC} = y</math>. Using law of cosines on <math>ABC</math> to find <math>AC</math> in terms of <math>y</math> and plugging that into the extended law of sines, we get <math>\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6</math>. Squaring both sides and cross multiplying, we get <math>36x^2 - 40x + 5 = 0</math>. Now, we get <math>x = \frac{10 \pm \sqrt{55}}{18}</math> using quadratic formula. If you drew a decent diagram, <math>B</math> is acute and therefore <math>x = \frac{10 + \sqrt{55}}{18}</math>(You can also try plugging in both in the end and seeing which gives a rational solution). Note that <math>BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.</math> Using the cosine addition formula and then plugging in what we know about <math>QBO</math>, we get that <math>BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}</math>. Now, the hard part is to find what <math>\sin{B}</math> is. We therefore want <math>\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}</math>. For the numerator, by inspection <math>(a + b\sqrt{55})^2</math> will not work for integers <math>a</math> and <math>b</math>. The other case is if there is <math>(a\sqrt{5} + b\sqrt{11})^2</math>. By inspection, <math>5\sqrt{5} - 2\sqrt{11}</math> works. Therefore, plugging all this in yields the answer, <math>\frac{18}{5} \rightarrow \boxed{23}</math>. Solution by hyxue<br />
<br />
===Solution 6===<br />
<asy> <br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(15cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.7673964645097335, xmax = 9.475267639476614, ymin = -1.6884766592324019, ymax = 6.385449160754665; /* image dimensions */<br />
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
/* draw figures */<br />
draw(circle((0.7129306199257198,2.4781596958650733), 3.000319171815248), linewidth(2) + wrwrwr); <br />
draw((0.7129306199257198,2.4781596958650733)--(3.178984115621537,0.7692140299269852), linewidth(2) + wrwrwr); <br />
draw((xmin, 1.4430262733614363*xmin + 1.4493820802284032)--(xmax, 1.4430262733614363*xmax + 1.4493820802284032), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, -0.020161290322580634*xmin + 0.8333064516129032)--(xmax, -0.020161290322580634*xmax + 0.8333064516129032), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, -8.047527437688247*xmin + 26.352175924366414)--(xmax, -8.047527437688247*xmax + 26.352175924366414), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, -2.5113572383524088*xmin + 8.752778799300463)--(xmax, -2.5113572383524088*xmax + 8.752778799300463), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, 0.12426176956126818*xmin + 2.389569675458691)--(xmax, 0.12426176956126818*xmax + 2.389569675458691), linewidth(2) + wrwrwr); /* line */<br />
draw(circle((1.9173376033752174,4.895608471162773), 0.7842529827808445), linewidth(2) + wrwrwr); <br />
/* dots and labels */<br />
dot((-1.82,0.87),dotstyle); <br />
label("$A$", (-1.7801363959463627,0.965838014692327), NE * labelscalefactor); <br />
dot((3.178984115621537,0.7692140299269852),dotstyle); <br />
label("$B$", (3.2140445236332655,0.8641046996638531), NE * labelscalefactor); <br />
dot((2.6857306099246263,4.738685150758791),dotstyle); <br />
label("$C$", (2.7238749148597092,4.831703985774336), NE * labelscalefactor); <br />
dot((0.7129306199257198,2.4781596958650733),linewidth(4pt) + dotstyle); <br />
label("$O$", (0.7539479965810783,2.556577122410283), NE * labelscalefactor); <br />
dot((-0.42105034508654754,0.8417953698606159),linewidth(4pt) + dotstyle); <br />
label("$P$", (-0.38361543510094825,0.9195955987702934), NE * labelscalefactor); <br />
dot((2.6239558409689123,5.235819298886746),linewidth(4pt) + dotstyle); <br />
label("$Q$", (2.6591355325688624,5.312625111363486), NE * labelscalefactor); <br />
dot((1.3292769824200672,5.414489427724579),linewidth(4pt) + dotstyle); <br />
label("$A'$", (1.3643478867519216,5.488346291867214), NE * labelscalefactor); <br />
dot((1.8469115849379867,4.11452402186953),linewidth(4pt) + dotstyle); <br />
label("$P'$", (1.8822629450786978,4.184310162865866), NE * labelscalefactor); <br />
dot((2.5624172335003985,5.731052930966743),linewidth(4pt) + dotstyle); <br />
label("$D$", (2.603644633462422,5.802794720137042), NE * labelscalefactor); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
</asy><br />
Reflect <math>A</math>, <math>P</math> across <math>OB</math> to points <math>A'</math> and <math>P'</math>, respectively with <math>A'</math> on the circle and <math>P, O, P'</math> collinear. Now, <math>\angle A'CQ = 180^{\circ} - \angle A'CB = \angle A'AB = \angle P'PB</math> by parallel lines. From here, <math>\angle P'PB = \angle PP'B = \angle A'P'Q</math> as <math>P, P', Q</math> collinear. From here, <math>A'P'QC</math> is cyclic, and by power of a point we obtain <math>\frac{18}{5} \implies \boxed{023}</math>.<br />
~awang11's sol<br />
<br />
==See also==<br />
{{AIME box|year=2015|n=II|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_14&diff=1196372019 AIME I Problems/Problem 142020-03-17T20:32:24Z<p>Hiker: /* Solution 2(basbhbashabsbhbasbasbhasbhsabhasbhbhasbhbash) */</p>
<hr />
<div>==Problem 14==<br />
Find the least odd prime factor of <math>2019^8+1</math>.<br />
<br />
==Solution==<br />
<br />
The problem tells us that <math>2019^8 \equiv -1 \pmod{p}</math> for some prime <math>p</math>. We want to find the smallest odd possible value of <math>p</math>. By squaring both sides of the congruence, we get <math>2019^{16} \equiv 1 \pmod{p}</math>. <br />
<br />
Since <math>2019^{16} \equiv 1 \pmod{p}</math>, <math>ord_p(2019)</math> = <math>1, 2, 4, 8,</math> or <math>16</math><br />
<br />
However, if <math>ord_p(2019)</math> = <math>1, 2, 4,</math> or <math>8,</math> then <math>2019^8</math> clearly will be <math>1 \pmod{p} </math> instead of <math>-1 \pmod{p}</math>, causing a contradiction.<br />
<br />
Therefore, <math>ord_p(2019) = 16</math>. Because <math>ord_p(2019) \vert \phi(p)</math>, <math>\phi(p)</math> is a multiple of 16. Since we know <math>p</math> is prime, <math>\phi(p) = p(1 - \frac{1}{p})</math> or <math>p - 1</math>. Therefore, <math>p</math> must be <math>1 \pmod{16}</math>. The two smallest primes that are <math>1 \pmod{16}</math> are <math>17</math> and <math>97</math>. <math>2019^8 \not\equiv -1 \pmod{17}</math>, but <math>2019^8 \equiv -1 \pmod{97}</math>, so our answer is <math>\boxed{97}</math>.<br />
<br />
===Note to solution===<br />
<math>\phi(p)</math> is called the [[Euler Totient Function]] of integer <math>p</math>.<br />
[[Euler's Totient Theorem]]: define <math>\phi(p)</math> as the number of positive integers less than <math>p</math> but relatively prime to <math>p</math>, then we have <cmath>\phi(p)=p\cdot \prod^n_{i=1}(1-\frac{1}{p_i})</cmath> where <math>p_1,p_2,...,p_n</math> are the prime factors of <math>p</math>. Then, we have <cmath>a^{\phi(p)} \equiv 1\ (\mathrm{mod}\ p)</cmath> if <math>\gcd(a,p)=1</math>.<br />
<br />
Furthermore, <math>ord_n(a)</math> for an integer <math>a</math> relatively prime to <math>n</math> is defined as the smallest positive integer <math>d</math> such that <math>a^{d} \equiv 1\ (\mathrm{mod}\ n)</math>. An important property of the order is that <math>ord_n(a)|\phi(n)</math>.<br />
<br />
==Video Solution==<br />
On The Spot STEM:<br />
<br />
https://youtu.be/_vHq5_5qCd8<br />
<br />
<br />
<br />
https://youtu.be/IF88iO5keFo<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_19&diff=1152822019 AMC 12A Problems/Problem 192020-01-23T21:57:46Z<p>Hiker: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
In <math>\triangle ABC</math> with integer side lengths,<br />
<cmath>\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \text{and} \qquad\cos C=-\frac{1}{4}.</cmath><br />
What is the least possible perimeter for <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math><br />
<br />
==Solution 1==<br />
Notice that by the Law of Sines, <math>a:b:c = \sin{A}:\sin{B}:\sin{C}</math>, so let's flip all the cosines using <math>\sin^{2}{x} + \cos^{2}{x} = 1</math> (<math>\sin{x}</math> is positive for <math>0^{\circ} < x < 180^{\circ}</math>, so we're good there).<br />
<br />
<math>\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}</math><br />
<br />
These are in the ratio <math>3:2:4</math>, so our minimal triangle has side lengths <math>2</math>, <math>3</math>, and <math>4</math>. <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
==Solution 2==<br />
<math>\angle ACB</math> is obtuse since its cosine is negative, so we let the foot of the altitude from <math>C</math> to <math>AB</math> be <math>H</math>. Let <math>AH=11x</math>, <math>AC=16x</math>, <math>BH=7y</math>, and <math>BC=8y</math>. By the Pythagorean Theorem, <math>CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}</math> and <math>CH=\sqrt{64y^2-49y^2}=y\sqrt{15}</math>. Thus, <math>y=3x</math>. The sides of the triangle are then <math>16x</math>, <math>11x+7(3x)=32x</math>, and <math>24x</math>, so for some integers <math>a,b</math>, <math>16x=a</math> and <math>24x=b</math>, where <math>a</math> and <math>b</math> are minimal. Hence, <math>\frac{a}{16}=\frac{b}{24}</math>, or <math>3a=2b</math>. Thus the smallest possible positive integers <math>a</math> and <math>b</math> that satisfy this are <math>a=2</math> and <math>b=3</math>, so <math>x=\frac{1}{8}</math>. The sides of the triangle are <math>2</math>, <math>3</math>, and <math>4</math>, so <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
<br />
==Solution 3==<br />
<br />
Using the law of cosines, we get the following equations:<br />
<br />
<cmath>c^2=a^2+b^2+\frac{ab}{2}</cmath><br />
<cmath>b^2=a^2+c^2-\frac{7ac}{4}</cmath><br />
<cmath>a^2=b^2+c^2-\frac{11bc}{8}</cmath><br />
<br />
Substituting <math>a^2+c^2-\frac{7ac}{4}</math> for <math>b^2</math> in <math>a^2=b^2+c^2-\frac{11bc}{8}</math> and simplifying, we get the following: <br />
<cmath>14a+11b=16c</cmath><br />
<br />
Note that since <math>a, b, c</math> are integers, we can solve this for integers. By some trial and error, we get that <math>(a,b,c) = (3,2,4)</math>. Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is <math>3+2+4 = \boxed{\textbf{(A) }9}</math>.<br />
<br />
~hiker<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2019|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_19&diff=1152812019 AMC 12A Problems/Problem 192020-01-23T21:57:36Z<p>Hiker: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
In <math>\triangle ABC</math> with integer side lengths,<br />
<cmath>\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \text{and} \qquad\cos C=-\frac{1}{4}.</cmath><br />
What is the least possible perimeter for <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math><br />
<br />
==Solution 1==<br />
Notice that by the Law of Sines, <math>a:b:c = \sin{A}:\sin{B}:\sin{C}</math>, so let's flip all the cosines using <math>\sin^{2}{x} + \cos^{2}{x} = 1</math> (\sin{x} is positive for <math>0^{\circ} < x < 180^{\circ}</math>, so we're good there).<br />
<br />
<math>\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}</math><br />
<br />
These are in the ratio <math>3:2:4</math>, so our minimal triangle has side lengths <math>2</math>, <math>3</math>, and <math>4</math>. <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
==Solution 2==<br />
<math>\angle ACB</math> is obtuse since its cosine is negative, so we let the foot of the altitude from <math>C</math> to <math>AB</math> be <math>H</math>. Let <math>AH=11x</math>, <math>AC=16x</math>, <math>BH=7y</math>, and <math>BC=8y</math>. By the Pythagorean Theorem, <math>CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}</math> and <math>CH=\sqrt{64y^2-49y^2}=y\sqrt{15}</math>. Thus, <math>y=3x</math>. The sides of the triangle are then <math>16x</math>, <math>11x+7(3x)=32x</math>, and <math>24x</math>, so for some integers <math>a,b</math>, <math>16x=a</math> and <math>24x=b</math>, where <math>a</math> and <math>b</math> are minimal. Hence, <math>\frac{a}{16}=\frac{b}{24}</math>, or <math>3a=2b</math>. Thus the smallest possible positive integers <math>a</math> and <math>b</math> that satisfy this are <math>a=2</math> and <math>b=3</math>, so <math>x=\frac{1}{8}</math>. The sides of the triangle are <math>2</math>, <math>3</math>, and <math>4</math>, so <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
<br />
==Solution 3==<br />
<br />
Using the law of cosines, we get the following equations:<br />
<br />
<cmath>c^2=a^2+b^2+\frac{ab}{2}</cmath><br />
<cmath>b^2=a^2+c^2-\frac{7ac}{4}</cmath><br />
<cmath>a^2=b^2+c^2-\frac{11bc}{8}</cmath><br />
<br />
Substituting <math>a^2+c^2-\frac{7ac}{4}</math> for <math>b^2</math> in <math>a^2=b^2+c^2-\frac{11bc}{8}</math> and simplifying, we get the following: <br />
<cmath>14a+11b=16c</cmath><br />
<br />
Note that since <math>a, b, c</math> are integers, we can solve this for integers. By some trial and error, we get that <math>(a,b,c) = (3,2,4)</math>. Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is <math>3+2+4 = \boxed{\textbf{(A) }9}</math>.<br />
<br />
~hiker<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2019|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_19&diff=1152802019 AMC 12A Problems/Problem 192020-01-23T21:57:16Z<p>Hiker: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
In <math>\triangle ABC</math> with integer side lengths,<br />
<cmath>\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \text{and} \qquad\cos C=-\frac{1}{4}.</cmath><br />
What is the least possible perimeter for <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math><br />
<br />
==Solution 1==<br />
Notice that by the Law of Sines, <math>a:b:c = \sin{A}:\sin{B}:\sin{C}</math>, so let's flip all the cosines using <math>\sin^{2}{x} + \cos^{2}{x} = 1</math> (sine is positive for <math>0^{\circ} < x < 180^{\circ}</math>, so we're good there).<br />
<br />
<math>\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}</math><br />
<br />
These are in the ratio <math>3:2:4</math>, so our minimal triangle has side lengths <math>2</math>, <math>3</math>, and <math>4</math>. <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
==Solution 2==<br />
<math>\angle ACB</math> is obtuse since its cosine is negative, so we let the foot of the altitude from <math>C</math> to <math>AB</math> be <math>H</math>. Let <math>AH=11x</math>, <math>AC=16x</math>, <math>BH=7y</math>, and <math>BC=8y</math>. By the Pythagorean Theorem, <math>CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}</math> and <math>CH=\sqrt{64y^2-49y^2}=y\sqrt{15}</math>. Thus, <math>y=3x</math>. The sides of the triangle are then <math>16x</math>, <math>11x+7(3x)=32x</math>, and <math>24x</math>, so for some integers <math>a,b</math>, <math>16x=a</math> and <math>24x=b</math>, where <math>a</math> and <math>b</math> are minimal. Hence, <math>\frac{a}{16}=\frac{b}{24}</math>, or <math>3a=2b</math>. Thus the smallest possible positive integers <math>a</math> and <math>b</math> that satisfy this are <math>a=2</math> and <math>b=3</math>, so <math>x=\frac{1}{8}</math>. The sides of the triangle are <math>2</math>, <math>3</math>, and <math>4</math>, so <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
<br />
==Solution 3==<br />
<br />
Using the law of cosines, we get the following equations:<br />
<br />
<cmath>c^2=a^2+b^2+\frac{ab}{2}</cmath><br />
<cmath>b^2=a^2+c^2-\frac{7ac}{4}</cmath><br />
<cmath>a^2=b^2+c^2-\frac{11bc}{8}</cmath><br />
<br />
Substituting <math>a^2+c^2-\frac{7ac}{4}</math> for <math>b^2</math> in <math>a^2=b^2+c^2-\frac{11bc}{8}</math> and simplifying, we get the following: <br />
<cmath>14a+11b=16c</cmath><br />
<br />
Note that since <math>a, b, c</math> are integers, we can solve this for integers. By some trial and error, we get that <math>(a,b,c) = (3,2,4)</math>. Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is <math>3+2+4 = \boxed{\textbf{(A) }9}</math>.<br />
<br />
~hiker<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2019|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_19&diff=1152792019 AMC 12A Problems/Problem 192020-01-23T21:56:00Z<p>Hiker: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
In <math>\triangle ABC</math> with integer side lengths,<br />
<cmath>\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \text{and} \qquad\cos C=-\frac{1}{4}.</cmath><br />
What is the least possible perimeter for <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math><br />
<br />
==Solution 1==<br />
Notice that by the Law of Sines, <math>a:b:c = \sin{A}:\sin{B}:\sin{C}</math>, so let's flip all the cosines using <math>\sin^{2}{x} + \cos^{2}{x} = 1</math> (sine is positive for <math>0^{\circ}\le x \le 180^{\circ}</math>, so we're good there).<br />
<br />
<math>\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}</math><br />
<br />
These are in the ratio <math>3:2:4</math>, so our minimal triangle has side lengths <math>2</math>, <math>3</math>, and <math>4</math>. <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
==Solution 2==<br />
<math>\angle ACB</math> is obtuse since its cosine is negative, so we let the foot of the altitude from <math>C</math> to <math>AB</math> be <math>H</math>. Let <math>AH=11x</math>, <math>AC=16x</math>, <math>BH=7y</math>, and <math>BC=8y</math>. By the Pythagorean Theorem, <math>CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}</math> and <math>CH=\sqrt{64y^2-49y^2}=y\sqrt{15}</math>. Thus, <math>y=3x</math>. The sides of the triangle are then <math>16x</math>, <math>11x+7(3x)=32x</math>, and <math>24x</math>, so for some integers <math>a,b</math>, <math>16x=a</math> and <math>24x=b</math>, where <math>a</math> and <math>b</math> are minimal. Hence, <math>\frac{a}{16}=\frac{b}{24}</math>, or <math>3a=2b</math>. Thus the smallest possible positive integers <math>a</math> and <math>b</math> that satisfy this are <math>a=2</math> and <math>b=3</math>, so <math>x=\frac{1}{8}</math>. The sides of the triangle are <math>2</math>, <math>3</math>, and <math>4</math>, so <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
<br />
==Solution 3==<br />
<br />
Using the law of cosines, we get the following equations:<br />
<br />
<cmath>c^2=a^2+b^2+\frac{ab}{2}</cmath><br />
<cmath>b^2=a^2+c^2-\frac{7ac}{4}</cmath><br />
<cmath>a^2=b^2+c^2-\frac{11bc}{8}</cmath><br />
<br />
Substituting <math>a^2+c^2-\frac{7ac}{4}</math> for <math>b^2</math> in <math>a^2=b^2+c^2-\frac{11bc}{8}</math> and simplifying, we get the following: <br />
<cmath>14a+11b=16c</cmath><br />
<br />
Note that since <math>a, b, c</math> are integers, we can solve this for integers. By some trial and error, we get that <math>(a,b,c) = (3,2,4)</math>. Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is <math>3+2+4 = \boxed{\textbf{(A) }9}</math>.<br />
<br />
~hiker<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2019|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems/Problem_7&diff=1135871995 AIME Problems/Problem 72019-12-28T16:38:08Z<p>Hiker: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Given that <math>(1+\sin t)(1+\cos t)=5/4</math> and<br />
:<math>(1-\sin t)(1-\cos t)=\frac mn-\sqrt{k},</math><br />
where <math>k, m,</math> and <math>n_{}</math> are [[positive integer]]s with <math>m_{}</math> and <math>n_{}</math> [[relatively prime]], find <math>k+m+n.</math><br />
<br />
== Solution ==<br />
From the givens, <br />
<math>2\sin t \cos t + 2 \sin t + 2 \cos t = \frac{1}{2}</math>, and adding <math>\sin^2 t + \cos^2t = 1</math> to both sides gives <math>(\sin t + \cos t)^2 + 2(\sin t + \cos t) = \frac{3}{2}</math>. Completing the square on the left in the variable <math>(\sin t + \cos t)</math> gives <math>\sin t + \cos t = -1 \pm \sqrt{\frac{5}{2}}</math>. Since <math>|\sin t + \cos t| \leq \sqrt 2 < 1 + \sqrt{\frac{5}{2}}</math>, we have <math>\sin t + \cos t = \sqrt{\frac{5}{2}} - 1</math>. Subtracting twice this from our original equation gives <math>(\sin t - 1)(\cos t - 1) = \sin t \cos t - \sin t - \cos t + 1 = \frac{13}{4} - \sqrt{10}</math>, so the answer is <math>13 + 4 + 10 = \boxed{027}</math>.<br />
<br />
== Solution 2 ==<br />
Let <math>(1 - \sin t)(1 - \cos t) = x</math>. Multiplying <math>x</math> with the given equation, <math>\frac{5x}{4} = (1 - \sin^2 t)(1 - \cos^2 t) = \sin^2 t \cos ^2 t</math>, and <math>\frac{\sqrt{5x}}{2} = \sin t \cos t</math>. Simplifying and rearranging the given equation, <math>\sin t + \cos t = \frac{5}{4} - (\sin^2 t + \cos^2 t) - \sin t \cos t = \frac{1}{4} - \frac{\sqrt{5x}}{2}</math>. Notice that <math>(1 + \sin t)(1 + \cos t) - 2(\sin t + \cos t) = x</math>, and substituting, <math>x = \frac{5}{4} - 2( \frac{1}{4} - \frac{\sqrt{5x}}{2}) = \frac{3}{4} + \sqrt{5x}</math>. Rearranging and squaring, <math>5x = x^2 - \frac{3}{2} x + \frac{9}{16}</math>, so <math>x^2 - \frac{13}{2} x + \frac{9}{16} = 0</math>, and <math>x = \frac{13}{4} \pm \sqrt{10}</math>, but clearly, <math>0 \leq x < 4</math>. Therefore, <math>x = \frac{13}{4} - \sqrt{10}</math>, and the answer is <math> 13 + 4 + 10 = \boxed{027}</math>.<br />
<br />
== Solution 3 == <br />
<br />
We want <math>1+\sin x \cos x-\sin x-\cos x</math>. However, note that we only need to find <math>\sin x+\cos x</math>.<br />
<br />
Let <math>y = \sin x+\cos x \rightarrow y^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2\sin x \cos x</math><br />
<br />
From this we have <math>\sin x \cos x = \frac{y^2-1}{2}</math> and <math>\sin x + \cos x = y</math><br />
<br />
Substituting, we have <math>2y^2+4y-3=0 \rightarrow y = \frac {-2 \pm \sqrt{10}}{2}</math><br />
<br />
<math>\frac{5}{4} - 2(\frac{-2+\sqrt{10}}{2}) = \frac{13}{4}-\sqrt{10} \rightarrow 13+10+4=\boxed{027}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1995|num-b=6|num-a=8}}<br />
<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_20&diff=1081612008 AMC 12A Problems/Problem 202019-08-02T20:31:19Z<p>Hiker: </p>
<hr />
<div>==Problem==<br />
Triangle <math>ABC</math> has <math>AC=3</math>, <math>BC=4</math>, and <math>AB=5</math>. Point <math>D</math> is on <math>\overline{AB}</math>, and <math>\overline{CD}</math> bisects the right angle. The inscribed circles of <math>\triangle ADC</math> and <math>\triangle BCD</math> have radii <math>r_a</math> and <math>r_b</math>, respectively. What is <math>r_a/r_b</math>?<br />
<br />
<math>\mathrm{(A)}\ \frac{1}{28}\left(10-\sqrt{2}\right)\qquad\mathrm{(B)}\ \frac{3}{56}\left(10-\sqrt{2}\right)\qquad\mathrm{(C)}\ \frac{1}{14}\left(10-\sqrt{2}\right)\qquad\mathrm{(D)}\ \frac{5}{56}\left(10-\sqrt{2}\right)\\\mathrm{(E)}\ \frac{3}{28}\left(10-\sqrt{2}\right)</math><br />
<br />
== Solution 1==<br />
<center><asy><br />
import olympiad;<br />
size(300);<br />
defaultpen(0.8);<br />
pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429);<br />
pair O=incenter(A,C,D), P=incenter(B,C,D);<br />
picture p = new picture; <br />
draw(p,Circle(C,0.2)); draw(p,Circle(B,0.2));<br />
clip(p,B--C--D--cycle);<br />
add(p);<br />
draw(A--B--C--D--C--cycle);<br />
draw(incircle(A,C,D));<br />
draw(incircle(B,C,D));<br />
dot(O);dot(P);<br />
label("\(A\)",A,W);<br />
label("\(B\)",B,E);<br />
label("\(C\)",C,W);<br />
label("\(D\)",D,NE);<br />
label("\(O_A\)",O,W);<br />
label("\(O_B\)",P,W);<br />
label("\(3\)",(A+C)/2,W);<br />
label("\(4\)",(B+C)/2,S);<br />
label("\(\frac{15}{7}\)",(A+D)/2,NE);<br />
label("\(\frac{20}{7}\)",(B+D)/2,NE);<br />
label("\(45^{\circ}\)",(.2,.1),E);<br />
label("\(\sin \theta = \frac{3}{5}\)",B-(.2,-.1),W);<br />
</asy></center><br />
<br />
By the [[Angle Bisector Theorem]], <br />
<cmath>\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7</cmath><br />
By [[Law of Sines]] on <math>\triangle BCD</math>, <br />
<cmath>\frac{BD}{\sin 45^{\circ}} = \frac{CD}{\sin \angle B} \Longrightarrow \frac{20/7}{\sqrt{2}/2} = \frac{CD}{3/5} \Longrightarrow CD=\frac{12\sqrt{2}}{7}</cmath><br />
Since the area of a triangle satisfies <math>[\triangle]=rs</math>, where <math>r = </math> the [[inradius]] and <math>s =</math> the [[semiperimeter]], we have <br />
<cmath>\frac{r_A}{r_B} = \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A}</cmath><br />
<!--Using any of various formulas for triangle area, we find the area <math>[BCD]</math> to be <br />
<cmath>[BCD] = \frac{1}{2} (\sin \angle CBD) \cdot (BD) \cdot (CD) = \frac 12 \cdot \frac 35 \cdot \frac{20}{7} \cdot 4 = \frac{24}{7}</cmath><br />
and <br />
<cmath>[ACD] = [ABC] - [BCD] = \frac 12 (3)(4) - \frac{24}{7} = \frac{18}{7}</cmath>--><br />
Since <math>\triangle ACD</math> and <math>\triangle BCD</math> share the [[altitude]] (to <math>\overline{AB}</math>), their areas are the ratio of their bases, or <cmath>\frac{[ACD]}{[BCD]} = \frac{AD}{BD} = \frac{3}{4}</cmath><br />
The semiperimeters are <math>s_A = \left(3 + \frac{15}{7} + \frac{12\sqrt{2}}{7}\right)\left/\right.2 = \frac{18+6\sqrt{2}}{7}</math> and <math>s_B = \frac{24+ 6\sqrt{2}}{7}</math>. Thus,<br />
<cmath>\begin{align*}<br />
\frac{r_A}{r_B} &= \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A} = \frac{3}{4} \cdot \frac{(24+ 6\sqrt{2})/7}{(18+6\sqrt{2})/7} \\<br />
&= \frac{3(4+\sqrt{2})}{4(3+\sqrt{2})} \cdot \left(\frac{3-\sqrt{2}}{3-\sqrt{2}}\right) = \frac{3}{28}(10-\sqrt{2}) \Rightarrow \mathrm{(E)}\qquad \blacksquare \end{align*}</cmath><br />
<br />
==Solution 2==<br />
<center><asy><br />
import olympiad;<br />
import geometry;<br />
size(300);<br />
defaultpen(0.8);<br />
<br />
pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429);<br />
pair O=incenter(A,C,D), P=incenter(B,C,D);<br />
line cd = line(C, D);<br />
<br />
picture p = new picture;<br />
picture q = new picture; <br />
picture r = new picture;<br />
picture s = new picture;<br />
<br />
draw(p,Circle(C,0.2));<br />
clip(p,P--C--D--cycle);<br />
<br />
draw(q, Circle(C, 0.3));<br />
clip(q, O--C--D--cycle);<br />
<br />
line l1 = perpendicular(O, cd);<br />
draw(r, l1);<br />
clip(r, C--D--O--cycle);<br />
<br />
line l2 = perpendicular(P, cd);<br />
draw(s, l2);<br />
clip(s, C--P--D--cycle); <br />
<br />
add(p);<br />
add(q);<br />
add(r);<br />
add(s);<br />
<br />
draw(A--B--C--D--C--cycle);<br />
draw(incircle(A,C,D));<br />
draw(incircle(B,C,D));<br />
draw(C--O);<br />
draw(C--P);<br />
dot(O);<br />
dot(P);<br />
<br />
point inter1 = intersectionpoint(l1, cd);<br />
point inter2 = intersectionpoint(l2, cd);<br />
dot(inter1);<br />
dot(inter2);<br />
<br />
label("\(A\)",A,W);<br />
label("\(B\)",B,E);<br />
label("\(C\)",C,W);<br />
label("\(D\)",D,NE);<br />
label("\(O_a\)",O,W);<br />
label("\(O_b\)",P,E);<br />
label("\(3\)",(A+C)/2,W);<br />
label("\(4\)",(B+C)/2,S);<br />
label("\(\frac{15}{7}\)",(A+D)/2,NE);<br />
label("\(\frac{20}{7}\)",(B+D)/2,NE);<br />
label("\(M\)", inter1, 2W);<br />
label("\(N\)", inter2, 2E);<br />
</asy></center><br />
<br />
We start by finding the length of <math>AD</math> and <math>BD</math> as in solution 1. Using the angle bisector theorem, we see that <math>AD = \frac{15}{7}</math> and <math>BD = \frac{20}{7}</math>. Using Stewart's Theorem gives us the equation <math>5d^2 + \frac{1500}{49} = \frac{240}{7} + \frac{180}{7}</math>, where <math>d</math> is the length of <math>CD</math>. Solving gives us <math>d = \frac{12\sqrt{2}}{7}</math>, so <math>CD = \frac{12\sqrt{2}}{7}</math>.<br />
<br />
Call the incenters of triangles <math>ACD</math> and <math>BCD</math> <math>O_a</math> and <math>O_b</math> respectively. Since <math>O_a</math> is an incenter, it lies on the angle bisector of <math>\angle ACD</math>. Similarly, <math>O_b</math> lies on the angle bisector of <math>\angle BCD</math>. Call the point on <math>CD</math> tangent to <math>O_a</math> <math>M</math>, and the point tangent to <math>O_b</math> <math>N</math>. Since <math>\triangle CO_aM</math> and <math>\triangle CO_bN</math> are right, and <math>\angle O_aCM = \angle O_bCN</math>, <math>\triangle CO_aM \sim \triangle CO_bN</math>. Then, <math>\frac{r_a}{r_b} = \frac{CM}{CN}</math>. <br />
<br />
We now use common tangents to find the length of <math>CM</math> and <math>CN</math>. Let <math>CM = m</math>, and the length of the other tangents be <math>n</math> and <math>p</math>. Since common tangents are equal, we can write that <math>m + n = \frac{12\sqrt{2}}{7}</math>, <math>n + p = \frac{15}{7}</math> and <math>m + p = 3</math>. Solving gives us that <math>CM = m = \frac{6\sqrt{2} + 3}{7}</math>. Similarly, <math>CN = \frac{6\sqrt{2} + 4}{7}</math>. <br />
<br />
We see now that <math>\frac{r_a}{r_b} = \frac{\frac{6\sqrt{2} + 3}{7}}{\frac{6\sqrt{2} + 4}{7}} = \frac{6\sqrt{2} + 3}{6\sqrt{2} + 4} = \frac{60-6\sqrt{2}}{56} = \frac{3}{28}(10 - \sqrt{2}) \Rightarrow \boxed{E}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|num-b=19|num-a=21|ab=A}}<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_14&diff=1081082005 AIME II Problems/Problem 142019-08-01T00:34:41Z<p>Hiker: /* Solution 7 (Super fast solution, 2 billion IQ) */</p>
<hr />
<div>== Problem ==<br />
<br />
In [[triangle]] <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such that <math> \angle BAE\cong \angle CAD. </math> Given that <math> BE=\frac pq </math> where <math> p </math> and <math> q </math> are relatively prime positive integers, find <math> q. </math><br />
<br />
== Solution 1==<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(150);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
</asy></center><br />
<br />
By the [[Law of Sines]] and since <math>\angle BAE = \angle CAD, \angle BAD = \angle CAE</math>, we have<br />
<br />
<cmath>\begin{align*}<br />
\frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\<br />
&= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\<br />
&= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2}<br />
\end{align*}<br />
</cmath><br />
<br />
Substituting our knowns, we have <math>\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}</math>. The answer is <math>q = \boxed{463}</math>.<br />
<br />
== Solution 2 (Similar Triangles)==<br />
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. <br />
<br />
From here, we can use Heron's Formula to find the altitude. The area of the triangle is <math>\sqrt{21*6*7*8} = 84</math>. We can then use similar triangles with triangle <math>AQC</math> and triangle <math>DSC</math> to find <math>DS=\frac{24}{5}</math>. Consequently, from Pythagorean theorem, <math>SC = \frac{18}{5}</math> and <math>AS = 14-SC = \frac{52}{5}</math>. We can also use the Pythagorean theorem on triangle <math>AQB</math> to determine that <math>BQ = \frac{33}{5}</math>.<br />
<br />
Label <math>AR</math> as <math>y</math> and <math>RE</math> as <math>x</math>. <math>RB</math> then equals <math>13-y</math>. Then, we have two similar triangles. <br />
<br />
Firstly: <math>\triangle ARE \sim \triangle ASD</math>. From there, we have <math>\frac{x}{y} = \frac{\frac{24}{5}}{\frac{53}{5}}</math>. <br />
<br />
Next: <math>\triangle BRE \sim \triangle BQA</math>. From there, we have <math>\frac{x}{13-y} = \frac{\frac{56}{5}}{\frac{33}{5}}</math>.<br />
<br />
Solve the system to get <math>x = \frac{2184}{463}</math> and <math>y = \frac{4732}{463}</math>. Notice that 463 is prime, so even though we use the Pythagorean theorem on <math>x</math> and <math>13-y</math>, the denominator won't change. The answer we desire is <math>\boxed{463}</math>.<br />
<br />
==Solution 3 (LoC and LoS bash)==<br />
Let <math>\angle CAD = \angle BAE = \theta</math>. Note by Law of Sines on <math>\triangle BEA</math> we have<br />
<cmath>\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{\angle BEA}}</cmath><br />
As a result, our goal is to find <math>\sin{\angle BEA}</math> and <math>\sin{\theta}</math> (we already know <math>AB</math>). <br />
<br />
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<br />
<cmath>169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}</cmath><br />
It follows that <math>AH = \frac{56}{5}</math> and <math>HC = \frac{42}{5} \Rightarrow AD = \frac{12}{5}</math>. <br />
<br />
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<br />
<cmath>\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}.</cmath><br />
How are we going to find <math>\sin{\angle BEA}</math> though? <math>\angle BEA</math> and <math>\theta</math> are in the same triangle. Applying Law of Sines on <math>\triangle ABC</math> we see that<br />
<cmath>\frac{13}{\frac{4}{5}} = \frac{14}{\sin{\angle B}} \Rightarrow \sin{\angle B} = \frac{56}{65} \Rightarrow \cos{\angle B} = \frac{33}{65}.</cmath><br />
<math>\theta</math>, <math>\angle B</math>, and <math>\angle BEA</math> are all in the same triangle. We know they add up to <math>180^{\circ}</math>. There's a good chance we can exploit this using the identity <math>\sin{p} = \sin{180^{\circ}-p}</math>.<br />
<br />
We have that <math>\sin{(180^{\circ} - (\theta + \angle B))} = \sin{\angle BEA} = \sin{(\theta + \angle B)}</math>. Success! We know <math>\sin{\theta}</math> and <math>\sin{\angle B}</math> already. Applying the <math>\sin</math> addition formula we see<br />
<cmath>\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}=\frac{1}{65 \cdot \sqrt{205}} (198 + 728) = \frac{926}{65 \sqrt{205}}.</cmath><br />
This is the last stretch! Applying Law of Sines a final time on <math>\triangle BEA</math> we see<br />
<cmath>\frac{BE}{\sin{\theta}} = \frac{13}{\sin{BEA}} \Rightarrow \frac{BE}{\frac{6}{\sqrt{205}}} = \frac{13}{\frac{926}{65\sqrt{205}}} \Rightarrow \frac{BE}{6} = \frac{13 \cdot 65}{926} \Rightarrow \frac{13 \cdot 65 \cdot 6}{926} = BE = \frac{2535}{463}.</cmath><br />
It follows that the answer is <math>\boxed{463}</math>.<br />
<br />
==Solution 4 (Ratio Lemma and Angle Bisector Theorem)==<br />
<br />
Let <math>AK</math> be the angle bisector of <math>\angle A</math> such that <math>K</math> is on <math>BC</math>.<br />
<br />
Then <math>\angle KAB = \angle KAC</math>, and thus <math>\angle KAE = \angle KAD</math>.<br />
<br />
By the Ratio Lemma, <br />
<math>\frac{BE}{KE} = \frac{BA}{KA} * \frac{\sin{BAE}}{\sin{KAE}}</math> and <math>\frac{CD}{KD} = \frac{CA}{KA} * \frac{\sin{CAD}}{\sin{KAD}}</math>.<br />
<br />
This implies that <math>\frac{BE}{KE*BA} = \frac{CD}{KD*CA}</math>.<br />
<br />
Thus, <math>\frac{BE}{KE} = \frac{13}{14} * \frac{6}{DK}</math>.<br />
<br />
<math>DK = CK - 6 = 14*15/27 - 6 = 16/9</math>. Thus, <math>\frac{BE}{KE} = \frac{13*54}{14*16}</math>.<br />
<br />
Additionally, <math>BE + KE = 9</math>. Solving gives that <math> q = 463.</math><br />
<br />
Alternate:<br />
By the ratio lemma,<br />
<math>BD/DC = (13/14)*(\sin BAD/\sin DAC)</math><br />
<math>EC/EB = (14/13)*(\sin EAC/\sin BAE)</math><br />
<br />
Combining these, we get<br />
<math>(BD/DC)(14/13) = (EC/EB)(13/14)</math><br />
<math>(3/2)(14/13)(14/13) = (15-x)(x)</math><br />
<br />
<math>x = 2535/463</math><br />
Thus, <math>q = 463</math><br />
<br />
==Solution 5 (Isogonal lines with respect to A angle bisesector)==<br />
Since <math>AE</math> and <math>AD</math> are isogonal with respect to the <math>A</math> angle bisector, we have <cmath>\frac{BE}{EC}\cdot \frac{BD}{DC}=(\frac{AB}{AC})^2.</cmath> To prove this, let <math>\angle BAE=\angle DAC=x</math> and <math>\angle BAD=\angle CAE=y.</math> Then, by the Ratio Lemma, we have <cmath> \frac{BD}{DC}=\frac{AB\sin y}{AC\sin x}</cmath> <cmath> \frac{BE}{EC}=\frac{AB\sin x}{AC\sin y}</cmath> and multiplying these together proves the formula for isogonal lines. Hence, we have <cmath>\frac{BE}{15-BE}\cdot \frac{9}{6}=\frac{169}{196}\implies BE=\frac{2535}{463}</cmath> so our desired answer is <math>\boxed{463}.</math><br />
<br />
<br />
<br />
==Solution 6 (Tangent subtraction formulas)==<br />
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. <br />
<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(300);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
pair G = foot(E,C,A);<br />
pair F = foot(D,C,A);<br />
draw(D--F);<br />
draw(E--G);<br />
label("$G$",G,N);<br />
label("$F$",F,N);<br />
</asy></center><br />
<br />
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{DG} = \frac{24}{5}</math> and <math>\overline{GC} = \frac{18}{5}</math>. Therefore, <math>\overline{AG} = \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>. <br />
<br />
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><br />
<br />
== Solution 7 (Super fast solution, 2 billion IQ)==<br />
<br />
Let <math><BAE = \angle CAB = \theta, \angle EAD = \alpha, BE = x \rightarrow ED = 9-x</math><br />
<br />
Via ratio lemma, we have<br />
<math>\frac{BD}{DC} = \frac{AB}{AC} \cdot \frac{\sin (\theta + \alpha)}{\sin (\theta)}</math> and <math>\frac{BE}{EC} = \frac{AB}{AC} \cdot \frac{\sin (\theta)}{\sin (\theta + \alpha)}</math><br />
<br />
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 = \frac{2535}{463} \rightarrow \boxed{463}</math><br />
<br />
Solution by hiker.<br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=II|num-b=13|num-a=15|t=368562}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_14&diff=1081072005 AIME II Problems/Problem 142019-08-01T00:34:16Z<p>Hiker: /* Solution 7 (Super fast solution, 2 billion IQ) */</p>
<hr />
<div>== Problem ==<br />
<br />
In [[triangle]] <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such that <math> \angle BAE\cong \angle CAD. </math> Given that <math> BE=\frac pq </math> where <math> p </math> and <math> q </math> are relatively prime positive integers, find <math> q. </math><br />
<br />
== Solution 1==<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(150);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
</asy></center><br />
<br />
By the [[Law of Sines]] and since <math>\angle BAE = \angle CAD, \angle BAD = \angle CAE</math>, we have<br />
<br />
<cmath>\begin{align*}<br />
\frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\<br />
&= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\<br />
&= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2}<br />
\end{align*}<br />
</cmath><br />
<br />
Substituting our knowns, we have <math>\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}</math>. The answer is <math>q = \boxed{463}</math>.<br />
<br />
== Solution 2 (Similar Triangles)==<br />
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. <br />
<br />
From here, we can use Heron's Formula to find the altitude. The area of the triangle is <math>\sqrt{21*6*7*8} = 84</math>. We can then use similar triangles with triangle <math>AQC</math> and triangle <math>DSC</math> to find <math>DS=\frac{24}{5}</math>. Consequently, from Pythagorean theorem, <math>SC = \frac{18}{5}</math> and <math>AS = 14-SC = \frac{52}{5}</math>. We can also use the Pythagorean theorem on triangle <math>AQB</math> to determine that <math>BQ = \frac{33}{5}</math>.<br />
<br />
Label <math>AR</math> as <math>y</math> and <math>RE</math> as <math>x</math>. <math>RB</math> then equals <math>13-y</math>. Then, we have two similar triangles. <br />
<br />
Firstly: <math>\triangle ARE \sim \triangle ASD</math>. From there, we have <math>\frac{x}{y} = \frac{\frac{24}{5}}{\frac{53}{5}}</math>. <br />
<br />
Next: <math>\triangle BRE \sim \triangle BQA</math>. From there, we have <math>\frac{x}{13-y} = \frac{\frac{56}{5}}{\frac{33}{5}}</math>.<br />
<br />
Solve the system to get <math>x = \frac{2184}{463}</math> and <math>y = \frac{4732}{463}</math>. Notice that 463 is prime, so even though we use the Pythagorean theorem on <math>x</math> and <math>13-y</math>, the denominator won't change. The answer we desire is <math>\boxed{463}</math>.<br />
<br />
==Solution 3 (LoC and LoS bash)==<br />
Let <math>\angle CAD = \angle BAE = \theta</math>. Note by Law of Sines on <math>\triangle BEA</math> we have<br />
<cmath>\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{\angle BEA}}</cmath><br />
As a result, our goal is to find <math>\sin{\angle BEA}</math> and <math>\sin{\theta}</math> (we already know <math>AB</math>). <br />
<br />
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<br />
<cmath>169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}</cmath><br />
It follows that <math>AH = \frac{56}{5}</math> and <math>HC = \frac{42}{5} \Rightarrow AD = \frac{12}{5}</math>. <br />
<br />
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<br />
<cmath>\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}.</cmath><br />
How are we going to find <math>\sin{\angle BEA}</math> though? <math>\angle BEA</math> and <math>\theta</math> are in the same triangle. Applying Law of Sines on <math>\triangle ABC</math> we see that<br />
<cmath>\frac{13}{\frac{4}{5}} = \frac{14}{\sin{\angle B}} \Rightarrow \sin{\angle B} = \frac{56}{65} \Rightarrow \cos{\angle B} = \frac{33}{65}.</cmath><br />
<math>\theta</math>, <math>\angle B</math>, and <math>\angle BEA</math> are all in the same triangle. We know they add up to <math>180^{\circ}</math>. There's a good chance we can exploit this using the identity <math>\sin{p} = \sin{180^{\circ}-p}</math>.<br />
<br />
We have that <math>\sin{(180^{\circ} - (\theta + \angle B))} = \sin{\angle BEA} = \sin{(\theta + \angle B)}</math>. Success! We know <math>\sin{\theta}</math> and <math>\sin{\angle B}</math> already. Applying the <math>\sin</math> addition formula we see<br />
<cmath>\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}=\frac{1}{65 \cdot \sqrt{205}} (198 + 728) = \frac{926}{65 \sqrt{205}}.</cmath><br />
This is the last stretch! Applying Law of Sines a final time on <math>\triangle BEA</math> we see<br />
<cmath>\frac{BE}{\sin{\theta}} = \frac{13}{\sin{BEA}} \Rightarrow \frac{BE}{\frac{6}{\sqrt{205}}} = \frac{13}{\frac{926}{65\sqrt{205}}} \Rightarrow \frac{BE}{6} = \frac{13 \cdot 65}{926} \Rightarrow \frac{13 \cdot 65 \cdot 6}{926} = BE = \frac{2535}{463}.</cmath><br />
It follows that the answer is <math>\boxed{463}</math>.<br />
<br />
==Solution 4 (Ratio Lemma and Angle Bisector Theorem)==<br />
<br />
Let <math>AK</math> be the angle bisector of <math>\angle A</math> such that <math>K</math> is on <math>BC</math>.<br />
<br />
Then <math>\angle KAB = \angle KAC</math>, and thus <math>\angle KAE = \angle KAD</math>.<br />
<br />
By the Ratio Lemma, <br />
<math>\frac{BE}{KE} = \frac{BA}{KA} * \frac{\sin{BAE}}{\sin{KAE}}</math> and <math>\frac{CD}{KD} = \frac{CA}{KA} * \frac{\sin{CAD}}{\sin{KAD}}</math>.<br />
<br />
This implies that <math>\frac{BE}{KE*BA} = \frac{CD}{KD*CA}</math>.<br />
<br />
Thus, <math>\frac{BE}{KE} = \frac{13}{14} * \frac{6}{DK}</math>.<br />
<br />
<math>DK = CK - 6 = 14*15/27 - 6 = 16/9</math>. Thus, <math>\frac{BE}{KE} = \frac{13*54}{14*16}</math>.<br />
<br />
Additionally, <math>BE + KE = 9</math>. Solving gives that <math> q = 463.</math><br />
<br />
Alternate:<br />
By the ratio lemma,<br />
<math>BD/DC = (13/14)*(\sin BAD/\sin DAC)</math><br />
<math>EC/EB = (14/13)*(\sin EAC/\sin BAE)</math><br />
<br />
Combining these, we get<br />
<math>(BD/DC)(14/13) = (EC/EB)(13/14)</math><br />
<math>(3/2)(14/13)(14/13) = (15-x)(x)</math><br />
<br />
<math>x = 2535/463</math><br />
Thus, <math>q = 463</math><br />
<br />
==Solution 5 (Isogonal lines with respect to A angle bisesector)==<br />
Since <math>AE</math> and <math>AD</math> are isogonal with respect to the <math>A</math> angle bisector, we have <cmath>\frac{BE}{EC}\cdot \frac{BD}{DC}=(\frac{AB}{AC})^2.</cmath> To prove this, let <math>\angle BAE=\angle DAC=x</math> and <math>\angle BAD=\angle CAE=y.</math> Then, by the Ratio Lemma, we have <cmath> \frac{BD}{DC}=\frac{AB\sin y}{AC\sin x}</cmath> <cmath> \frac{BE}{EC}=\frac{AB\sin x}{AC\sin y}</cmath> and multiplying these together proves the formula for isogonal lines. Hence, we have <cmath>\frac{BE}{15-BE}\cdot \frac{9}{6}=\frac{169}{196}\implies BE=\frac{2535}{463}</cmath> so our desired answer is <math>\boxed{463}.</math><br />
<br />
<br />
<br />
==Solution 6 (Tangent subtraction formulas)==<br />
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. <br />
<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(300);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
pair G = foot(E,C,A);<br />
pair F = foot(D,C,A);<br />
draw(D--F);<br />
draw(E--G);<br />
label("$G$",G,N);<br />
label("$F$",F,N);<br />
</asy></center><br />
<br />
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{DG} = \frac{24}{5}</math> and <math>\overline{GC} = \frac{18}{5}</math>. Therefore, <math>\overline{AG} = \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>. <br />
<br />
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><br />
<br />
== Solution 7 (Super fast solution, 2 billion IQ)==<br />
<br />
Let <math><BAE = \angle CAB = \theta, \angle EAD = \alpha, BE = x \rightarrow ED = 9-x</math><br />
<br />
Via ratio lemma, we have<br />
<math>\frac{BD}{DC} = \frac{AB}{AC} \cdot \frac{\sin (\theta + \alpha)}{\sin (\theta)}</math> and <math>\frac{BE}{EC} = \frac{AB}{AC} \cdot \frac{\sin (\theta)}{\sin (\theta + \alpha)}</math><br />
<br />
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 = \frac{2535}{463} \rightarrow \boxed{463}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=II|num-b=13|num-a=15|t=368562}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_14&diff=1081062005 AIME II Problems/Problem 142019-08-01T00:33:42Z<p>Hiker: /* Solution 7 (Super fast solution, 2 billion IQ) */</p>
<hr />
<div>== Problem ==<br />
<br />
In [[triangle]] <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such that <math> \angle BAE\cong \angle CAD. </math> Given that <math> BE=\frac pq </math> where <math> p </math> and <math> q </math> are relatively prime positive integers, find <math> q. </math><br />
<br />
== Solution 1==<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(150);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
</asy></center><br />
<br />
By the [[Law of Sines]] and since <math>\angle BAE = \angle CAD, \angle BAD = \angle CAE</math>, we have<br />
<br />
<cmath>\begin{align*}<br />
\frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\<br />
&= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\<br />
&= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2}<br />
\end{align*}<br />
</cmath><br />
<br />
Substituting our knowns, we have <math>\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}</math>. The answer is <math>q = \boxed{463}</math>.<br />
<br />
== Solution 2 (Similar Triangles)==<br />
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. <br />
<br />
From here, we can use Heron's Formula to find the altitude. The area of the triangle is <math>\sqrt{21*6*7*8} = 84</math>. We can then use similar triangles with triangle <math>AQC</math> and triangle <math>DSC</math> to find <math>DS=\frac{24}{5}</math>. Consequently, from Pythagorean theorem, <math>SC = \frac{18}{5}</math> and <math>AS = 14-SC = \frac{52}{5}</math>. We can also use the Pythagorean theorem on triangle <math>AQB</math> to determine that <math>BQ = \frac{33}{5}</math>.<br />
<br />
Label <math>AR</math> as <math>y</math> and <math>RE</math> as <math>x</math>. <math>RB</math> then equals <math>13-y</math>. Then, we have two similar triangles. <br />
<br />
Firstly: <math>\triangle ARE \sim \triangle ASD</math>. From there, we have <math>\frac{x}{y} = \frac{\frac{24}{5}}{\frac{53}{5}}</math>. <br />
<br />
Next: <math>\triangle BRE \sim \triangle BQA</math>. From there, we have <math>\frac{x}{13-y} = \frac{\frac{56}{5}}{\frac{33}{5}}</math>.<br />
<br />
Solve the system to get <math>x = \frac{2184}{463}</math> and <math>y = \frac{4732}{463}</math>. Notice that 463 is prime, so even though we use the Pythagorean theorem on <math>x</math> and <math>13-y</math>, the denominator won't change. The answer we desire is <math>\boxed{463}</math>.<br />
<br />
==Solution 3 (LoC and LoS bash)==<br />
Let <math>\angle CAD = \angle BAE = \theta</math>. Note by Law of Sines on <math>\triangle BEA</math> we have<br />
<cmath>\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{\angle BEA}}</cmath><br />
As a result, our goal is to find <math>\sin{\angle BEA}</math> and <math>\sin{\theta}</math> (we already know <math>AB</math>). <br />
<br />
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<br />
<cmath>169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}</cmath><br />
It follows that <math>AH = \frac{56}{5}</math> and <math>HC = \frac{42}{5} \Rightarrow AD = \frac{12}{5}</math>. <br />
<br />
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<br />
<cmath>\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}.</cmath><br />
How are we going to find <math>\sin{\angle BEA}</math> though? <math>\angle BEA</math> and <math>\theta</math> are in the same triangle. Applying Law of Sines on <math>\triangle ABC</math> we see that<br />
<cmath>\frac{13}{\frac{4}{5}} = \frac{14}{\sin{\angle B}} \Rightarrow \sin{\angle B} = \frac{56}{65} \Rightarrow \cos{\angle B} = \frac{33}{65}.</cmath><br />
<math>\theta</math>, <math>\angle B</math>, and <math>\angle BEA</math> are all in the same triangle. We know they add up to <math>180^{\circ}</math>. There's a good chance we can exploit this using the identity <math>\sin{p} = \sin{180^{\circ}-p}</math>.<br />
<br />
We have that <math>\sin{(180^{\circ} - (\theta + \angle B))} = \sin{\angle BEA} = \sin{(\theta + \angle B)}</math>. Success! We know <math>\sin{\theta}</math> and <math>\sin{\angle B}</math> already. Applying the <math>\sin</math> addition formula we see<br />
<cmath>\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}=\frac{1}{65 \cdot \sqrt{205}} (198 + 728) = \frac{926}{65 \sqrt{205}}.</cmath><br />
This is the last stretch! Applying Law of Sines a final time on <math>\triangle BEA</math> we see<br />
<cmath>\frac{BE}{\sin{\theta}} = \frac{13}{\sin{BEA}} \Rightarrow \frac{BE}{\frac{6}{\sqrt{205}}} = \frac{13}{\frac{926}{65\sqrt{205}}} \Rightarrow \frac{BE}{6} = \frac{13 \cdot 65}{926} \Rightarrow \frac{13 \cdot 65 \cdot 6}{926} = BE = \frac{2535}{463}.</cmath><br />
It follows that the answer is <math>\boxed{463}</math>.<br />
<br />
==Solution 4 (Ratio Lemma and Angle Bisector Theorem)==<br />
<br />
Let <math>AK</math> be the angle bisector of <math>\angle A</math> such that <math>K</math> is on <math>BC</math>.<br />
<br />
Then <math>\angle KAB = \angle KAC</math>, and thus <math>\angle KAE = \angle KAD</math>.<br />
<br />
By the Ratio Lemma, <br />
<math>\frac{BE}{KE} = \frac{BA}{KA} * \frac{\sin{BAE}}{\sin{KAE}}</math> and <math>\frac{CD}{KD} = \frac{CA}{KA} * \frac{\sin{CAD}}{\sin{KAD}}</math>.<br />
<br />
This implies that <math>\frac{BE}{KE*BA} = \frac{CD}{KD*CA}</math>.<br />
<br />
Thus, <math>\frac{BE}{KE} = \frac{13}{14} * \frac{6}{DK}</math>.<br />
<br />
<math>DK = CK - 6 = 14*15/27 - 6 = 16/9</math>. Thus, <math>\frac{BE}{KE} = \frac{13*54}{14*16}</math>.<br />
<br />
Additionally, <math>BE + KE = 9</math>. Solving gives that <math> q = 463.</math><br />
<br />
Alternate:<br />
By the ratio lemma,<br />
<math>BD/DC = (13/14)*(\sin BAD/\sin DAC)</math><br />
<math>EC/EB = (14/13)*(\sin EAC/\sin BAE)</math><br />
<br />
Combining these, we get<br />
<math>(BD/DC)(14/13) = (EC/EB)(13/14)</math><br />
<math>(3/2)(14/13)(14/13) = (15-x)(x)</math><br />
<br />
<math>x = 2535/463</math><br />
Thus, <math>q = 463</math><br />
<br />
==Solution 5 (Isogonal lines with respect to A angle bisesector)==<br />
Since <math>AE</math> and <math>AD</math> are isogonal with respect to the <math>A</math> angle bisector, we have <cmath>\frac{BE}{EC}\cdot \frac{BD}{DC}=(\frac{AB}{AC})^2.</cmath> To prove this, let <math>\angle BAE=\angle DAC=x</math> and <math>\angle BAD=\angle CAE=y.</math> Then, by the Ratio Lemma, we have <cmath> \frac{BD}{DC}=\frac{AB\sin y}{AC\sin x}</cmath> <cmath> \frac{BE}{EC}=\frac{AB\sin x}{AC\sin y}</cmath> and multiplying these together proves the formula for isogonal lines. Hence, we have <cmath>\frac{BE}{15-BE}\cdot \frac{9}{6}=\frac{169}{196}\implies BE=\frac{2535}{463}</cmath> so our desired answer is <math>\boxed{463}.</math><br />
<br />
<br />
<br />
==Solution 6 (Tangent subtraction formulas)==<br />
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. <br />
<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(300);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
pair G = foot(E,C,A);<br />
pair F = foot(D,C,A);<br />
draw(D--F);<br />
draw(E--G);<br />
label("$G$",G,N);<br />
label("$F$",F,N);<br />
</asy></center><br />
<br />
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{DG} = \frac{24}{5}</math> and <math>\overline{GC} = \frac{18}{5}</math>. Therefore, <math>\overline{AG} = \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>. <br />
<br />
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><br />
<br />
== Solution 7 (Super fast solution, 2 billion IQ)==<br />
<br />
Let <math><BAE = <CAB = \theta, <EAD = \alpha, BE = x \rightarrow ED = 9-x</math><br />
<br />
Via ratio lemma, we have<br />
<math>\frac{BD}{DC} = \frac{AB}{AC} \cdot \frac{\sin (\theta + \alpha)}{\sin (\theta)}</math> and <math>\frac{BE}{EC} = \frac{AB}{AC} \cdot \frac{\sin (\theta)}{\sin (\theta + \alpha)}</math><br />
<br />
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 = \frac{2535}{463} \rightarrow \boxed{463}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=II|num-b=13|num-a=15|t=368562}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_14&diff=1081052005 AIME II Problems/Problem 142019-08-01T00:17:08Z<p>Hiker: /* Solution 7 (Super fast solution, 2 billion IQ) */</p>
<hr />
<div>== Problem ==<br />
<br />
In [[triangle]] <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such that <math> \angle BAE\cong \angle CAD. </math> Given that <math> BE=\frac pq </math> where <math> p </math> and <math> q </math> are relatively prime positive integers, find <math> q. </math><br />
<br />
== Solution 1==<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(150);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
</asy></center><br />
<br />
By the [[Law of Sines]] and since <math>\angle BAE = \angle CAD, \angle BAD = \angle CAE</math>, we have<br />
<br />
<cmath>\begin{align*}<br />
\frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\<br />
&= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\<br />
&= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2}<br />
\end{align*}<br />
</cmath><br />
<br />
Substituting our knowns, we have <math>\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}</math>. The answer is <math>q = \boxed{463}</math>.<br />
<br />
== Solution 2 (Similar Triangles)==<br />
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. <br />
<br />
From here, we can use Heron's Formula to find the altitude. The area of the triangle is <math>\sqrt{21*6*7*8} = 84</math>. We can then use similar triangles with triangle <math>AQC</math> and triangle <math>DSC</math> to find <math>DS=\frac{24}{5}</math>. Consequently, from Pythagorean theorem, <math>SC = \frac{18}{5}</math> and <math>AS = 14-SC = \frac{52}{5}</math>. We can also use the Pythagorean theorem on triangle <math>AQB</math> to determine that <math>BQ = \frac{33}{5}</math>.<br />
<br />
Label <math>AR</math> as <math>y</math> and <math>RE</math> as <math>x</math>. <math>RB</math> then equals <math>13-y</math>. Then, we have two similar triangles. <br />
<br />
Firstly: <math>\triangle ARE \sim \triangle ASD</math>. From there, we have <math>\frac{x}{y} = \frac{\frac{24}{5}}{\frac{53}{5}}</math>. <br />
<br />
Next: <math>\triangle BRE \sim \triangle BQA</math>. From there, we have <math>\frac{x}{13-y} = \frac{\frac{56}{5}}{\frac{33}{5}}</math>.<br />
<br />
Solve the system to get <math>x = \frac{2184}{463}</math> and <math>y = \frac{4732}{463}</math>. Notice that 463 is prime, so even though we use the Pythagorean theorem on <math>x</math> and <math>13-y</math>, the denominator won't change. The answer we desire is <math>\boxed{463}</math>.<br />
<br />
==Solution 3 (LoC and LoS bash)==<br />
Let <math>\angle CAD = \angle BAE = \theta</math>. Note by Law of Sines on <math>\triangle BEA</math> we have<br />
<cmath>\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{\angle BEA}}</cmath><br />
As a result, our goal is to find <math>\sin{\angle BEA}</math> and <math>\sin{\theta}</math> (we already know <math>AB</math>). <br />
<br />
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<br />
<cmath>169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}</cmath><br />
It follows that <math>AH = \frac{56}{5}</math> and <math>HC = \frac{42}{5} \Rightarrow AD = \frac{12}{5}</math>. <br />
<br />
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<br />
<cmath>\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}.</cmath><br />
How are we going to find <math>\sin{\angle BEA}</math> though? <math>\angle BEA</math> and <math>\theta</math> are in the same triangle. Applying Law of Sines on <math>\triangle ABC</math> we see that<br />
<cmath>\frac{13}{\frac{4}{5}} = \frac{14}{\sin{\angle B}} \Rightarrow \sin{\angle B} = \frac{56}{65} \Rightarrow \cos{\angle B} = \frac{33}{65}.</cmath><br />
<math>\theta</math>, <math>\angle B</math>, and <math>\angle BEA</math> are all in the same triangle. We know they add up to <math>180^{\circ}</math>. There's a good chance we can exploit this using the identity <math>\sin{p} = \sin{180^{\circ}-p}</math>.<br />
<br />
We have that <math>\sin{(180^{\circ} - (\theta + \angle B))} = \sin{\angle BEA} = \sin{(\theta + \angle B)}</math>. Success! We know <math>\sin{\theta}</math> and <math>\sin{\angle B}</math> already. Applying the <math>\sin</math> addition formula we see<br />
<cmath>\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}=\frac{1}{65 \cdot \sqrt{205}} (198 + 728) = \frac{926}{65 \sqrt{205}}.</cmath><br />
This is the last stretch! Applying Law of Sines a final time on <math>\triangle BEA</math> we see<br />
<cmath>\frac{BE}{\sin{\theta}} = \frac{13}{\sin{BEA}} \Rightarrow \frac{BE}{\frac{6}{\sqrt{205}}} = \frac{13}{\frac{926}{65\sqrt{205}}} \Rightarrow \frac{BE}{6} = \frac{13 \cdot 65}{926} \Rightarrow \frac{13 \cdot 65 \cdot 6}{926} = BE = \frac{2535}{463}.</cmath><br />
It follows that the answer is <math>\boxed{463}</math>.<br />
<br />
==Solution 4 (Ratio Lemma and Angle Bisector Theorem)==<br />
<br />
Let <math>AK</math> be the angle bisector of <math>\angle A</math> such that <math>K</math> is on <math>BC</math>.<br />
<br />
Then <math>\angle KAB = \angle KAC</math>, and thus <math>\angle KAE = \angle KAD</math>.<br />
<br />
By the Ratio Lemma, <br />
<math>\frac{BE}{KE} = \frac{BA}{KA} * \frac{\sin{BAE}}{\sin{KAE}}</math> and <math>\frac{CD}{KD} = \frac{CA}{KA} * \frac{\sin{CAD}}{\sin{KAD}}</math>.<br />
<br />
This implies that <math>\frac{BE}{KE*BA} = \frac{CD}{KD*CA}</math>.<br />
<br />
Thus, <math>\frac{BE}{KE} = \frac{13}{14} * \frac{6}{DK}</math>.<br />
<br />
<math>DK = CK - 6 = 14*15/27 - 6 = 16/9</math>. Thus, <math>\frac{BE}{KE} = \frac{13*54}{14*16}</math>.<br />
<br />
Additionally, <math>BE + KE = 9</math>. Solving gives that <math> q = 463.</math><br />
<br />
Alternate:<br />
By the ratio lemma,<br />
<math>BD/DC = (13/14)*(\sin BAD/\sin DAC)</math><br />
<math>EC/EB = (14/13)*(\sin EAC/\sin BAE)</math><br />
<br />
Combining these, we get<br />
<math>(BD/DC)(14/13) = (EC/EB)(13/14)</math><br />
<math>(3/2)(14/13)(14/13) = (15-x)(x)</math><br />
<br />
<math>x = 2535/463</math><br />
Thus, <math>q = 463</math><br />
<br />
==Solution 5 (Isogonal lines with respect to A angle bisesector)==<br />
Since <math>AE</math> and <math>AD</math> are isogonal with respect to the <math>A</math> angle bisector, we have <cmath>\frac{BE}{EC}\cdot \frac{BD}{DC}=(\frac{AB}{AC})^2.</cmath> To prove this, let <math>\angle BAE=\angle DAC=x</math> and <math>\angle BAD=\angle CAE=y.</math> Then, by the Ratio Lemma, we have <cmath> \frac{BD}{DC}=\frac{AB\sin y}{AC\sin x}</cmath> <cmath> \frac{BE}{EC}=\frac{AB\sin x}{AC\sin y}</cmath> and multiplying these together proves the formula for isogonal lines. Hence, we have <cmath>\frac{BE}{15-BE}\cdot \frac{9}{6}=\frac{169}{196}\implies BE=\frac{2535}{463}</cmath> so our desired answer is <math>\boxed{463}.</math><br />
<br />
<br />
<br />
==Solution 6 (Tangent subtraction formulas)==<br />
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. <br />
<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(300);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
pair G = foot(E,C,A);<br />
pair F = foot(D,C,A);<br />
draw(D--F);<br />
draw(E--G);<br />
label("$G$",G,N);<br />
label("$F$",F,N);<br />
</asy></center><br />
<br />
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{DG} = \frac{24}{5}</math> and <math>\overline{GC} = \frac{18}{5}</math>. Therefore, <math>\overline{AG} = \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>. <br />
<br />
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><br />
<br />
== Solution 7 (Super fast solution, 2 billion IQ)==<br />
<br />
Let <math><BAE = <CAB = \theta, <EAD = \alpha, BE = x \rightarrow ED = 9-x</math><br />
<br />
Using Ratio Lemma 2 times, we obtain the following equations:<br />
<math><br />
\tfrac{BD}{DC} = \tfrac{AB}{AC} \cdot \tfrac{\sin (\theta + \alpha)}{\sin (\theta)}<br />
\tfrac{BE}{EC} = \tfrac{AB}{AC} \cdot \tfrac{\sin (\theta)}{\sin (\theta + \alpha)}</math><br />
<br />
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><br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=II|num-b=13|num-a=15|t=368562}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_14&diff=1081042005 AIME II Problems/Problem 142019-08-01T00:16:52Z<p>Hiker: /* Solution 7 (Super fast solution, 2 billion IQ) */</p>
<hr />
<div>== Problem ==<br />
<br />
In [[triangle]] <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such that <math> \angle BAE\cong \angle CAD. </math> Given that <math> BE=\frac pq </math> where <math> p </math> and <math> q </math> are relatively prime positive integers, find <math> q. </math><br />
<br />
== Solution 1==<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(150);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
</asy></center><br />
<br />
By the [[Law of Sines]] and since <math>\angle BAE = \angle CAD, \angle BAD = \angle CAE</math>, we have<br />
<br />
<cmath>\begin{align*}<br />
\frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\<br />
&= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\<br />
&= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2}<br />
\end{align*}<br />
</cmath><br />
<br />
Substituting our knowns, we have <math>\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}</math>. The answer is <math>q = \boxed{463}</math>.<br />
<br />
== Solution 2 (Similar Triangles)==<br />
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. <br />
<br />
From here, we can use Heron's Formula to find the altitude. The area of the triangle is <math>\sqrt{21*6*7*8} = 84</math>. We can then use similar triangles with triangle <math>AQC</math> and triangle <math>DSC</math> to find <math>DS=\frac{24}{5}</math>. Consequently, from Pythagorean theorem, <math>SC = \frac{18}{5}</math> and <math>AS = 14-SC = \frac{52}{5}</math>. We can also use the Pythagorean theorem on triangle <math>AQB</math> to determine that <math>BQ = \frac{33}{5}</math>.<br />
<br />
Label <math>AR</math> as <math>y</math> and <math>RE</math> as <math>x</math>. <math>RB</math> then equals <math>13-y</math>. Then, we have two similar triangles. <br />
<br />
Firstly: <math>\triangle ARE \sim \triangle ASD</math>. From there, we have <math>\frac{x}{y} = \frac{\frac{24}{5}}{\frac{53}{5}}</math>. <br />
<br />
Next: <math>\triangle BRE \sim \triangle BQA</math>. From there, we have <math>\frac{x}{13-y} = \frac{\frac{56}{5}}{\frac{33}{5}}</math>.<br />
<br />
Solve the system to get <math>x = \frac{2184}{463}</math> and <math>y = \frac{4732}{463}</math>. Notice that 463 is prime, so even though we use the Pythagorean theorem on <math>x</math> and <math>13-y</math>, the denominator won't change. The answer we desire is <math>\boxed{463}</math>.<br />
<br />
==Solution 3 (LoC and LoS bash)==<br />
Let <math>\angle CAD = \angle BAE = \theta</math>. Note by Law of Sines on <math>\triangle BEA</math> we have<br />
<cmath>\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{\angle BEA}}</cmath><br />
As a result, our goal is to find <math>\sin{\angle BEA}</math> and <math>\sin{\theta}</math> (we already know <math>AB</math>). <br />
<br />
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<br />
<cmath>169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}</cmath><br />
It follows that <math>AH = \frac{56}{5}</math> and <math>HC = \frac{42}{5} \Rightarrow AD = \frac{12}{5}</math>. <br />
<br />
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<br />
<cmath>\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}.</cmath><br />
How are we going to find <math>\sin{\angle BEA}</math> though? <math>\angle BEA</math> and <math>\theta</math> are in the same triangle. Applying Law of Sines on <math>\triangle ABC</math> we see that<br />
<cmath>\frac{13}{\frac{4}{5}} = \frac{14}{\sin{\angle B}} \Rightarrow \sin{\angle B} = \frac{56}{65} \Rightarrow \cos{\angle B} = \frac{33}{65}.</cmath><br />
<math>\theta</math>, <math>\angle B</math>, and <math>\angle BEA</math> are all in the same triangle. We know they add up to <math>180^{\circ}</math>. There's a good chance we can exploit this using the identity <math>\sin{p} = \sin{180^{\circ}-p}</math>.<br />
<br />
We have that <math>\sin{(180^{\circ} - (\theta + \angle B))} = \sin{\angle BEA} = \sin{(\theta + \angle B)}</math>. Success! We know <math>\sin{\theta}</math> and <math>\sin{\angle B}</math> already. Applying the <math>\sin</math> addition formula we see<br />
<cmath>\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}=\frac{1}{65 \cdot \sqrt{205}} (198 + 728) = \frac{926}{65 \sqrt{205}}.</cmath><br />
This is the last stretch! Applying Law of Sines a final time on <math>\triangle BEA</math> we see<br />
<cmath>\frac{BE}{\sin{\theta}} = \frac{13}{\sin{BEA}} \Rightarrow \frac{BE}{\frac{6}{\sqrt{205}}} = \frac{13}{\frac{926}{65\sqrt{205}}} \Rightarrow \frac{BE}{6} = \frac{13 \cdot 65}{926} \Rightarrow \frac{13 \cdot 65 \cdot 6}{926} = BE = \frac{2535}{463}.</cmath><br />
It follows that the answer is <math>\boxed{463}</math>.<br />
<br />
==Solution 4 (Ratio Lemma and Angle Bisector Theorem)==<br />
<br />
Let <math>AK</math> be the angle bisector of <math>\angle A</math> such that <math>K</math> is on <math>BC</math>.<br />
<br />
Then <math>\angle KAB = \angle KAC</math>, and thus <math>\angle KAE = \angle KAD</math>.<br />
<br />
By the Ratio Lemma, <br />
<math>\frac{BE}{KE} = \frac{BA}{KA} * \frac{\sin{BAE}}{\sin{KAE}}</math> and <math>\frac{CD}{KD} = \frac{CA}{KA} * \frac{\sin{CAD}}{\sin{KAD}}</math>.<br />
<br />
This implies that <math>\frac{BE}{KE*BA} = \frac{CD}{KD*CA}</math>.<br />
<br />
Thus, <math>\frac{BE}{KE} = \frac{13}{14} * \frac{6}{DK}</math>.<br />
<br />
<math>DK = CK - 6 = 14*15/27 - 6 = 16/9</math>. Thus, <math>\frac{BE}{KE} = \frac{13*54}{14*16}</math>.<br />
<br />
Additionally, <math>BE + KE = 9</math>. Solving gives that <math> q = 463.</math><br />
<br />
Alternate:<br />
By the ratio lemma,<br />
<math>BD/DC = (13/14)*(\sin BAD/\sin DAC)</math><br />
<math>EC/EB = (14/13)*(\sin EAC/\sin BAE)</math><br />
<br />
Combining these, we get<br />
<math>(BD/DC)(14/13) = (EC/EB)(13/14)</math><br />
<math>(3/2)(14/13)(14/13) = (15-x)(x)</math><br />
<br />
<math>x = 2535/463</math><br />
Thus, <math>q = 463</math><br />
<br />
==Solution 5 (Isogonal lines with respect to A angle bisesector)==<br />
Since <math>AE</math> and <math>AD</math> are isogonal with respect to the <math>A</math> angle bisector, we have <cmath>\frac{BE}{EC}\cdot \frac{BD}{DC}=(\frac{AB}{AC})^2.</cmath> To prove this, let <math>\angle BAE=\angle DAC=x</math> and <math>\angle BAD=\angle CAE=y.</math> Then, by the Ratio Lemma, we have <cmath> \frac{BD}{DC}=\frac{AB\sin y}{AC\sin x}</cmath> <cmath> \frac{BE}{EC}=\frac{AB\sin x}{AC\sin y}</cmath> and multiplying these together proves the formula for isogonal lines. Hence, we have <cmath>\frac{BE}{15-BE}\cdot \frac{9}{6}=\frac{169}{196}\implies BE=\frac{2535}{463}</cmath> so our desired answer is <math>\boxed{463}.</math><br />
<br />
<br />
<br />
==Solution 6 (Tangent subtraction formulas)==<br />
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. <br />
<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(300);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
pair G = foot(E,C,A);<br />
pair F = foot(D,C,A);<br />
draw(D--F);<br />
draw(E--G);<br />
label("$G$",G,N);<br />
label("$F$",F,N);<br />
</asy></center><br />
<br />
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{DG} = \frac{24}{5}</math> and <math>\overline{GC} = \frac{18}{5}</math>. Therefore, <math>\overline{AG} = \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>. <br />
<br />
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><br />
<br />
== Solution 7 (Super fast solution, 2 billion IQ)==<br />
<br />
Let <math><BAE = <CAB = \theta, <EAD = \alpha, BE = x \rightarrow ED = 9-x</math><br />
<br />
Using Ratio Lemma 2 times, we obtain the following equations:<br />
<math><br />
\tfrac{BD}{DC} = \tfrac{AB}{AC} \cdot \tfrac{\sin (\theta + \alpha)}{\sin (\theta)}<br />
<br />
2. \tfrac{BE}{EC} = \tfrac{AB}{AC} \cdot \tfrac{\sin (\theta)}{\sin (\theta + \alpha)}</math><br />
<br />
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><br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=II|num-b=13|num-a=15|t=368562}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_14&diff=1081032005 AIME II Problems/Problem 142019-08-01T00:16:27Z<p>Hiker: /* Solution 6 (Tangent subtraction formulas) */</p>
<hr />
<div>== Problem ==<br />
<br />
In [[triangle]] <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such that <math> \angle BAE\cong \angle CAD. </math> Given that <math> BE=\frac pq </math> where <math> p </math> and <math> q </math> are relatively prime positive integers, find <math> q. </math><br />
<br />
== Solution 1==<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(150);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
</asy></center><br />
<br />
By the [[Law of Sines]] and since <math>\angle BAE = \angle CAD, \angle BAD = \angle CAE</math>, we have<br />
<br />
<cmath>\begin{align*}<br />
\frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\<br />
&= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\<br />
&= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2}<br />
\end{align*}<br />
</cmath><br />
<br />
Substituting our knowns, we have <math>\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}</math>. The answer is <math>q = \boxed{463}</math>.<br />
<br />
== Solution 2 (Similar Triangles)==<br />
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. <br />
<br />
From here, we can use Heron's Formula to find the altitude. The area of the triangle is <math>\sqrt{21*6*7*8} = 84</math>. We can then use similar triangles with triangle <math>AQC</math> and triangle <math>DSC</math> to find <math>DS=\frac{24}{5}</math>. Consequently, from Pythagorean theorem, <math>SC = \frac{18}{5}</math> and <math>AS = 14-SC = \frac{52}{5}</math>. We can also use the Pythagorean theorem on triangle <math>AQB</math> to determine that <math>BQ = \frac{33}{5}</math>.<br />
<br />
Label <math>AR</math> as <math>y</math> and <math>RE</math> as <math>x</math>. <math>RB</math> then equals <math>13-y</math>. Then, we have two similar triangles. <br />
<br />
Firstly: <math>\triangle ARE \sim \triangle ASD</math>. From there, we have <math>\frac{x}{y} = \frac{\frac{24}{5}}{\frac{53}{5}}</math>. <br />
<br />
Next: <math>\triangle BRE \sim \triangle BQA</math>. From there, we have <math>\frac{x}{13-y} = \frac{\frac{56}{5}}{\frac{33}{5}}</math>.<br />
<br />
Solve the system to get <math>x = \frac{2184}{463}</math> and <math>y = \frac{4732}{463}</math>. Notice that 463 is prime, so even though we use the Pythagorean theorem on <math>x</math> and <math>13-y</math>, the denominator won't change. The answer we desire is <math>\boxed{463}</math>.<br />
<br />
==Solution 3 (LoC and LoS bash)==<br />
Let <math>\angle CAD = \angle BAE = \theta</math>. Note by Law of Sines on <math>\triangle BEA</math> we have<br />
<cmath>\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{\angle BEA}}</cmath><br />
As a result, our goal is to find <math>\sin{\angle BEA}</math> and <math>\sin{\theta}</math> (we already know <math>AB</math>). <br />
<br />
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<br />
<cmath>169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}</cmath><br />
It follows that <math>AH = \frac{56}{5}</math> and <math>HC = \frac{42}{5} \Rightarrow AD = \frac{12}{5}</math>. <br />
<br />
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<br />
<cmath>\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}.</cmath><br />
How are we going to find <math>\sin{\angle BEA}</math> though? <math>\angle BEA</math> and <math>\theta</math> are in the same triangle. Applying Law of Sines on <math>\triangle ABC</math> we see that<br />
<cmath>\frac{13}{\frac{4}{5}} = \frac{14}{\sin{\angle B}} \Rightarrow \sin{\angle B} = \frac{56}{65} \Rightarrow \cos{\angle B} = \frac{33}{65}.</cmath><br />
<math>\theta</math>, <math>\angle B</math>, and <math>\angle BEA</math> are all in the same triangle. We know they add up to <math>180^{\circ}</math>. There's a good chance we can exploit this using the identity <math>\sin{p} = \sin{180^{\circ}-p}</math>.<br />
<br />
We have that <math>\sin{(180^{\circ} - (\theta + \angle B))} = \sin{\angle BEA} = \sin{(\theta + \angle B)}</math>. Success! We know <math>\sin{\theta}</math> and <math>\sin{\angle B}</math> already. Applying the <math>\sin</math> addition formula we see<br />
<cmath>\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}=\frac{1}{65 \cdot \sqrt{205}} (198 + 728) = \frac{926}{65 \sqrt{205}}.</cmath><br />
This is the last stretch! Applying Law of Sines a final time on <math>\triangle BEA</math> we see<br />
<cmath>\frac{BE}{\sin{\theta}} = \frac{13}{\sin{BEA}} \Rightarrow \frac{BE}{\frac{6}{\sqrt{205}}} = \frac{13}{\frac{926}{65\sqrt{205}}} \Rightarrow \frac{BE}{6} = \frac{13 \cdot 65}{926} \Rightarrow \frac{13 \cdot 65 \cdot 6}{926} = BE = \frac{2535}{463}.</cmath><br />
It follows that the answer is <math>\boxed{463}</math>.<br />
<br />
==Solution 4 (Ratio Lemma and Angle Bisector Theorem)==<br />
<br />
Let <math>AK</math> be the angle bisector of <math>\angle A</math> such that <math>K</math> is on <math>BC</math>.<br />
<br />
Then <math>\angle KAB = \angle KAC</math>, and thus <math>\angle KAE = \angle KAD</math>.<br />
<br />
By the Ratio Lemma, <br />
<math>\frac{BE}{KE} = \frac{BA}{KA} * \frac{\sin{BAE}}{\sin{KAE}}</math> and <math>\frac{CD}{KD} = \frac{CA}{KA} * \frac{\sin{CAD}}{\sin{KAD}}</math>.<br />
<br />
This implies that <math>\frac{BE}{KE*BA} = \frac{CD}{KD*CA}</math>.<br />
<br />
Thus, <math>\frac{BE}{KE} = \frac{13}{14} * \frac{6}{DK}</math>.<br />
<br />
<math>DK = CK - 6 = 14*15/27 - 6 = 16/9</math>. Thus, <math>\frac{BE}{KE} = \frac{13*54}{14*16}</math>.<br />
<br />
Additionally, <math>BE + KE = 9</math>. Solving gives that <math> q = 463.</math><br />
<br />
Alternate:<br />
By the ratio lemma,<br />
<math>BD/DC = (13/14)*(\sin BAD/\sin DAC)</math><br />
<math>EC/EB = (14/13)*(\sin EAC/\sin BAE)</math><br />
<br />
Combining these, we get<br />
<math>(BD/DC)(14/13) = (EC/EB)(13/14)</math><br />
<math>(3/2)(14/13)(14/13) = (15-x)(x)</math><br />
<br />
<math>x = 2535/463</math><br />
Thus, <math>q = 463</math><br />
<br />
==Solution 5 (Isogonal lines with respect to A angle bisesector)==<br />
Since <math>AE</math> and <math>AD</math> are isogonal with respect to the <math>A</math> angle bisector, we have <cmath>\frac{BE}{EC}\cdot \frac{BD}{DC}=(\frac{AB}{AC})^2.</cmath> To prove this, let <math>\angle BAE=\angle DAC=x</math> and <math>\angle BAD=\angle CAE=y.</math> Then, by the Ratio Lemma, we have <cmath> \frac{BD}{DC}=\frac{AB\sin y}{AC\sin x}</cmath> <cmath> \frac{BE}{EC}=\frac{AB\sin x}{AC\sin y}</cmath> and multiplying these together proves the formula for isogonal lines. Hence, we have <cmath>\frac{BE}{15-BE}\cdot \frac{9}{6}=\frac{169}{196}\implies BE=\frac{2535}{463}</cmath> so our desired answer is <math>\boxed{463}.</math><br />
<br />
<br />
<br />
==Solution 6 (Tangent subtraction formulas)==<br />
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. <br />
<br />
<center><asy><br />
import olympiad; import cse5; import geometry; size(300);<br />
defaultpen(fontsize(10pt));<br />
defaultpen(0.8);<br />
dotfactor = 4;<br />
pair A = origin;<br />
pair C = rotate(15,A)*(A+dir(-50));<br />
pair B = rotate(15,A)*(A+dir(-130));<br />
pair D = extension(A,A+dir(-68),B,C);<br />
pair E = extension(A,A+dir(-82),B,C);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$D$",D,SE);<br />
label("$E$",E,S);<br />
label("$C$",C,SE);<br />
draw(A--B--C--cycle);<br />
draw(A--E);<br />
draw(A--D);<br />
draw(anglemark(B,A,E,5));<br />
draw(anglemark(D,A,C,5));<br />
pair G = foot(E,C,A);<br />
pair F = foot(D,C,A);<br />
draw(D--F);<br />
draw(E--G);<br />
label("$G$",G,N);<br />
label("$F$",F,N);<br />
</asy></center><br />
<br />
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{DG} = \frac{24}{5}</math> and <math>\overline{GC} = \frac{18}{5}</math>. Therefore, <math>\overline{AG} = \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>. <br />
<br />
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><br />
<br />
== Solution 7 (Super fast solution, 2 billion IQ)==<br />
<br />
Let <math><BAE = <CAB = \theta, <EAD = \alpha, BE = x \rightarrow ED = 9-x</math><br />
<br />
Using Ratio Lemma 2 times, we obtain the following equations:<br />
<math><br />
1. \tfrac{BD}{DC} = \tfrac{AB}{AC} \cdot \tfrac{\sin (\theta + \alpha)}{\sin (\theta) }<br />
2. \tfrac{BE}{EC} = \tfrac{AB}{AC} \cdot \tfrac{\sin (\theta)}{\sin (\theta + \alpha)}</math><br />
<br />
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><br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=II|num-b=13|num-a=15|t=368562}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems/Problem_7&diff=1080431995 AIME Problems/Problem 72019-07-30T16:42:54Z<p>Hiker: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Given that <math>(1+\sin t)(1+\cos t)=5/4</math> and<br />
:<math>(1-\sin t)(1-\cos t)=\frac mn-\sqrt{k},</math><br />
where <math>k, m,</math> and <math>n_{}</math> are [[positive integer]]s with <math>m_{}</math> and <math>n_{}</math> [[relatively prime]], find <math>k+m+n.</math><br />
<br />
== Solution ==<br />
From the givens, <br />
<math>2\sin t \cos t + 2 \sin t + 2 \cos t = \frac{1}{2}</math>, and adding <math>\sin^2 t + \cos^2t = 1</math> to both sides gives <math>(\sin t + \cos t)^2 + 2(\sin t + \cos t) = \frac{3}{2}</math>. Completing the square on the left in the variable <math>(\sin t + \cos t)</math> gives <math>\sin t + \cos t = -1 \pm \sqrt{\frac{5}{2}}</math>. Since <math>|\sin t + \cos t| \leq \sqrt 2 < 1 + \sqrt{\frac{5}{2}}</math>, we have <math>\sin t + \cos t = \sqrt{\frac{5}{2}} - 1</math>. Subtracting twice this from our original equation gives <math>(\sin t - 1)(\cos t - 1) = \sin t \cos t - \sin t - \cos t + 1 = \frac{13}{4} - \sqrt{10}</math>, so the answer is <math>13 + 4 + 10 = \boxed{027}</math>.<br />
<br />
== Solution 2 ==<br />
Let <math>(1 - \sin t)(1 - \cos t) = x</math>. Multiplying <math>x</math> with the given equation, <math>\frac{5x}{4} = (1 - \sin^2 t)(1 - \cos^2 t) = \sin^2 t \cos ^2 t</math>, and <math>\frac{\sqrt{5x}}{2} = \sin t \cos t</math>. Simplifying and rearranging the given equation, <math>\sin t + \cos t = \frac{5}{4} - (\sin^2 t + \cos^2 t) - \sin t \cos t = \frac{1}{4} - \frac{\sqrt{5x}}{2}</math>. Notice that <math>(1 + \sin t)(1 + \cos t) - 2(\sin t + \cos t) = x</math>, and substituting, <math>x = \frac{5}{4} - 2( \frac{1}{4} - \frac{\sqrt{5x}}{2}) = \frac{3}{4} + \sqrt{5x}</math>. Rearranging and squaring, <math>5x = x^2 - \frac{3}{2} x + \frac{9}{16}</math>, so <math>x^2 - \frac{13}{2} x + \frac{9}{16} = 0</math>, and <math>x = \frac{13}{4} \pm \sqrt{10}</math>, but clearly, <math>0 \leq x < 4</math>. Therefore, <math>x = \frac{13}{4} - \sqrt{10}</math>, and the answer is <math> 13 + 4 + 10 = \boxed{027}</math>.<br />
<br />
== Solution 3 == <br />
<br />
We would like to find <math>1+\sin x \cos x-\sin x-\cos x</math>. If we find <math>\sin x+\cos x</math>, we'll be done with the problem.<br />
<br />
Let <math>y = \sin x+\cos x \rightarrow y^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2\sin x \cos x</math><br />
<br />
From this we have <math>\sin x \cos x = \frac{y^2-1}{2}</math> and <math>\sin x + \cos x = y</math><br />
<br />
Substituting this into <math>1+\sin x\cos x+\sin x+\cos x = \frac{5}{4}</math>, we have <math>2y^2+4y-3=0 \rightarrow y = \frac {-2 \pm \sqrt{10}}{2}</math><br />
<br />
<math>\frac{5}{4} - 2(\frac{-2+\sqrt{10}}{2}) = \frac{13}{4}-\sqrt{10} \rightarrow 13+10+4=\boxed{027}</math>.<br />
<br />
(by hiker)<br />
<br />
== See also ==<br />
{{AIME box|year=1995|num-b=6|num-a=8}}<br />
<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=1996_AIME_Problems/Problem_10&diff=1080171996 AIME Problems/Problem 102019-07-29T18:41:07Z<p>Hiker: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Find the smallest positive integer solution to <math>\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}</math>.<br />
<br />
== Solution ==<br />
<math>\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} =</math> <math>\dfrac{\sin{186^{\circ}}+\sin{96^{\circ}}}{\sin{186^{\circ}}-\sin{96^{\circ}}} =</math> <math>\dfrac{\sin{(141^{\circ}+45^{\circ})}+\sin{(141^{\circ}-45^{\circ})}}{\sin{(141^{\circ}+45^{\circ})}-\sin{(141^{\circ}-45^{\circ})}} =</math> <math>\dfrac{2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}</math>. <br />
<br />
The period of the [[Trigonometry#Tangent|tangent]] function is <math>180^\circ</math>, and the tangent function is [[one-to-one]] over each period of its domain. <br />
<br />
Thus, <math>19x \equiv 141 \pmod{180}</math>. <br />
<br />
Since <math>19^2 \equiv 361 \equiv 1 \pmod{180}</math>, multiplying both sides by <math>19</math> yields <math>x \equiv 141 \cdot 19 \equiv (140+1)(18+1) \equiv 0+140+18+1 \equiv 159 \pmod{180}</math>. <br />
<br />
Therefore, the smallest positive solution is <math>x = \boxed{159}</math>.<br />
== Solution 2 ==<br />
<math>\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} = \dfrac{1 + \tan{96^{\circ}}}{1-\tan{96^{\circ}}}</math><br />
which is the same as <math>\dfrac{\tan{45^{\circ}} + \tan{96^{\circ}}}{1-\tan{45^{\circ}}\tan{96^{\circ}}} = \tan{141{^\circ}}</math>.<br />
<br />
So <math>19x = 141 +180n</math>, for some integer <math>n</math>.<br />
Multiplying by <math>19</math> gives <math>x \equiv 141 \cdot 19 \equiv 2679 \equiv 159 \pmod{180}</math>.<br />
The smallest positive solution of this is <math>x = \boxed{159}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1996|num-b=9|num-a=11}}<br />
<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_19&diff=1068502019 AMC 12A Problems/Problem 192019-06-23T02:30:08Z<p>Hiker: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
In <math>\triangle ABC</math> with integer side lengths,<br />
<cmath>\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \text{and} \qquad\cos C=-\frac{1}{4}.</cmath><br />
What is the least possible perimeter for <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math><br />
<br />
==Solution 1==<br />
Notice that by the Law of Sines, <math>a:b:c = \sin{A}:\sin{B}:\sin{C}</math>, so let's flip all the cosines using <math>\sin^{2}{x} + \cos^{2}{x} = 1</math> (sine is positive for <math>0^{\circ}\le x \le 180^{\circ}</math>, so we're good there).<br />
<br />
<math>\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}</math><br />
<br />
These are in the ratio <math>3:2:4</math>, so our minimal triangle has side lengths <math>2</math>, <math>3</math>, and <math>4</math>. <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
==Solution 2==<br />
<math>\angle ACB</math> is obtuse since its cosine is negative, so we let the foot of the altitude from <math>C</math> to <math>AB</math> be <math>H</math>. Let <math>AH=11x</math>, <math>AC=16x</math>, <math>BH=7y</math>, and <math>BC=8y</math>. By the Pythagorean Theorem, <math>CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}</math> and <math>CH=\sqrt{64y^2-49y^2}=y\sqrt{15}</math>. Thus, <math>y=3x</math>. The sides of the triangle are then <math>16x</math>, <math>11x+7(3x)=32x</math>, and <math>24x</math>, so for some integers <math>a,b</math>, <math>16x=a</math> and <math>24x=b</math>, where <math>a</math> and <math>b</math> are minimal. Hence, <math>\frac{a}{16}=\frac{b}{24}</math>, or <math>3a=2b</math>. Thus the smallest possible positive integers <math>a</math> and <math>b</math> that satisfy this are <math>a=2</math> and <math>b=3</math>, so <math>x=\frac{1}{8}</math>. The sides of the triangle are <math>2</math>, <math>3</math>, and <math>4</math>, so <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
<br />
==Solution 3==<br />
<br />
Using the law of cosines, we get the following equations:<br />
<br />
<cmath>c^2=a^2+b^2+\frac{ab}{2}</cmath><br />
<cmath>b^2=a^2+c^2-\frac{7ac}{4}</cmath><br />
<cmath>a^2=b^2+c^2-\frac{11bc}{8}</cmath><br />
<br />
Substituting <math>a^2+c^2-\frac{7ac}{4}</math> for <math>b^2</math> in <math>a^2=b^2+c^2-\frac{11bc}{8}</math> and simplifying, we get the following: <br />
<cmath>14a+11b+16c</cmath><br />
<br />
Note that since <math>a, b, c</math> are integers, we can solve this for integers. By some trial and error, we get that <math>(a,b,c) = (3,2,4)</math>. Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is <math>3+2+4 = \boxed{\textbf{(A) }9}</math>.<br />
<br />
Solution by hiker.<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2019|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_19&diff=1068492019 AMC 12A Problems/Problem 192019-06-23T02:29:54Z<p>Hiker: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
In <math>\triangle ABC</math> with integer side lengths,<br />
<cmath>\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \text{and} \qquad\cos C=-\frac{1}{4}.</cmath><br />
What is the least possible perimeter for <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math><br />
<br />
==Solution 1==<br />
Notice that by the Law of Sines, <math>a:b:c = \sin{A}:\sin{B}:\sin{C}</math>, so let's flip all the cosines using <math>\sin^{2}{x} + \cos^{2}{x} = 1</math> (sine is positive for <math>0^{\circ}\le x \le 180^{\circ}</math>, so we're good there).<br />
<br />
<math>\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}</math><br />
<br />
These are in the ratio <math>3:2:4</math>, so our minimal triangle has side lengths <math>2</math>, <math>3</math>, and <math>4</math>. <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
==Solution 2==<br />
<math>\angle ACB</math> is obtuse since its cosine is negative, so we let the foot of the altitude from <math>C</math> to <math>AB</math> be <math>H</math>. Let <math>AH=11x</math>, <math>AC=16x</math>, <math>BH=7y</math>, and <math>BC=8y</math>. By the Pythagorean Theorem, <math>CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}</math> and <math>CH=\sqrt{64y^2-49y^2}=y\sqrt{15}</math>. Thus, <math>y=3x</math>. The sides of the triangle are then <math>16x</math>, <math>11x+7(3x)=32x</math>, and <math>24x</math>, so for some integers <math>a,b</math>, <math>16x=a</math> and <math>24x=b</math>, where <math>a</math> and <math>b</math> are minimal. Hence, <math>\frac{a}{16}=\frac{b}{24}</math>, or <math>3a=2b</math>. Thus the smallest possible positive integers <math>a</math> and <math>b</math> that satisfy this are <math>a=2</math> and <math>b=3</math>, so <math>x=\frac{1}{8}</math>. The sides of the triangle are <math>2</math>, <math>3</math>, and <math>4</math>, so <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
<br />
==Solution 3==<br />
<br />
Using the law of cosines, we get the following equations:<br />
<br />
<cmath>c^2=a^2+b^2+\frac{ab}{2}</cmath><br />
<cmath>b^2=a^2+c^2-\frac{7ac}{4}</cmath><br />
<cmath>a^2=b^2+c^2-\frac{11bc}{8}</cmath><br />
<br />
Substituting <math>a^2+c^2-\frac{7ac}{4}</math> for <math>b^2</math> in <math>a^2=b^2+c^2-\frac{11bc}{8}</math> and simplifying, we get the following: <br />
<cmath>14a+11b+16c</cmath><br />
<br />
Note that since <math>a, b, c</math> are integers, we can solve this for integers. By some trial and error, we get that <math>(a,b,c) = (3,2,4)</math>. Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is <math>3+2+4 = \boxed{\textbf{(A)} 9}</math>.<br />
<br />
Solution by hiker.<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2019|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_19&diff=1068482019 AMC 12A Problems/Problem 192019-06-23T02:29:25Z<p>Hiker: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
In <math>\triangle ABC</math> with integer side lengths,<br />
<cmath>\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \text{and} \qquad\cos C=-\frac{1}{4}.</cmath><br />
What is the least possible perimeter for <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math><br />
<br />
==Solution 1==<br />
Notice that by the Law of Sines, <math>a:b:c = \sin{A}:\sin{B}:\sin{C}</math>, so let's flip all the cosines using <math>\sin^{2}{x} + \cos^{2}{x} = 1</math> (sine is positive for <math>0^{\circ}\le x \le 180^{\circ}</math>, so we're good there).<br />
<br />
<math>\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}</math><br />
<br />
These are in the ratio <math>3:2:4</math>, so our minimal triangle has side lengths <math>2</math>, <math>3</math>, and <math>4</math>. <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
==Solution 2==<br />
<math>\angle ACB</math> is obtuse since its cosine is negative, so we let the foot of the altitude from <math>C</math> to <math>AB</math> be <math>H</math>. Let <math>AH=11x</math>, <math>AC=16x</math>, <math>BH=7y</math>, and <math>BC=8y</math>. By the Pythagorean Theorem, <math>CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}</math> and <math>CH=\sqrt{64y^2-49y^2}=y\sqrt{15}</math>. Thus, <math>y=3x</math>. The sides of the triangle are then <math>16x</math>, <math>11x+7(3x)=32x</math>, and <math>24x</math>, so for some integers <math>a,b</math>, <math>16x=a</math> and <math>24x=b</math>, where <math>a</math> and <math>b</math> are minimal. Hence, <math>\frac{a}{16}=\frac{b}{24}</math>, or <math>3a=2b</math>. Thus the smallest possible positive integers <math>a</math> and <math>b</math> that satisfy this are <math>a=2</math> and <math>b=3</math>, so <math>x=\frac{1}{8}</math>. The sides of the triangle are <math>2</math>, <math>3</math>, and <math>4</math>, so <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
<br />
==Solution 3==<br />
<br />
Using the law of cosines, we get the following equations:<br />
<br />
<cmath>c^2=a^2+b^2+\frac{ab}{2}</cmath><br />
<cmath>b^2=a^2+c^2-\frac{7ac}{4}</cmath><br />
<cmath>a^2=b^2+c^2-\frac{11bc}{8}</cmath><br />
<br />
Substituting <math>a^2+c^2-\frac{7ac}{4}</math> for <math>b^2</math> in <math>a^2=b^2+c^2-\frac{11bc}{8}</math> and simplifying, we get the following: <br />
<cmath>14a+11b+16c</cmath><br />
<br />
Note that since <math>a, b, c</math> are integers, we can solve this for integers. By some trial and error, we get that <math>(a,b,c) = (3,2,4)</math>. Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is <math>3+2+4 = \boxed{9}</math>.<br />
<br />
Solution by hiker.<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2019|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_19&diff=1068472019 AMC 12A Problems/Problem 192019-06-23T02:29:12Z<p>Hiker: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
In <math>\triangle ABC</math> with integer side lengths,<br />
<cmath>\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \text{and} \qquad\cos C=-\frac{1}{4}.</cmath><br />
What is the least possible perimeter for <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math><br />
<br />
==Solution 1==<br />
Notice that by the Law of Sines, <math>a:b:c = \sin{A}:\sin{B}:\sin{C}</math>, so let's flip all the cosines using <math>\sin^{2}{x} + \cos^{2}{x} = 1</math> (sine is positive for <math>0^{\circ}\le x \le 180^{\circ}</math>, so we're good there).<br />
<br />
<math>\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}</math><br />
<br />
These are in the ratio <math>3:2:4</math>, so our minimal triangle has side lengths <math>2</math>, <math>3</math>, and <math>4</math>. <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
==Solution 2==<br />
<math>\angle ACB</math> is obtuse since its cosine is negative, so we let the foot of the altitude from <math>C</math> to <math>AB</math> be <math>H</math>. Let <math>AH=11x</math>, <math>AC=16x</math>, <math>BH=7y</math>, and <math>BC=8y</math>. By the Pythagorean Theorem, <math>CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}</math> and <math>CH=\sqrt{64y^2-49y^2}=y\sqrt{15}</math>. Thus, <math>y=3x</math>. The sides of the triangle are then <math>16x</math>, <math>11x+7(3x)=32x</math>, and <math>24x</math>, so for some integers <math>a,b</math>, <math>16x=a</math> and <math>24x=b</math>, where <math>a</math> and <math>b</math> are minimal. Hence, <math>\frac{a}{16}=\frac{b}{24}</math>, or <math>3a=2b</math>. Thus the smallest possible positive integers <math>a</math> and <math>b</math> that satisfy this are <math>a=2</math> and <math>b=3</math>, so <math>x=\frac{1}{8}</math>. The sides of the triangle are <math>2</math>, <math>3</math>, and <math>4</math>, so <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
<br />
==Solution 3==<br />
<br />
Using the law of cosines, we get the following equations:<br />
<br />
<cmath>c^2=a^2+b^2+\frac{ab}{2}</cmath><br />
<cmath>b^2=a^2+c^2-\frac{7ac}{4}</cmath><br />
<cmath>a^2=b^2+c^2-\frac{11bc}{8}</cmath><br />
<br />
Substituting <math>a^2+c^2-\frac{7ac}{4}</math> for <math>b^2</math> in <math>a^2=b^2+c^2-\frac{11bc}{8}</math> and simplifying, we get the following: <br />
<cmath>14a+11b+16c</cmath><br />
<br />
Note that since <math>a, b, c</math> are integers, we can solve this for integers. By some trial and error, we get that <math>(a,b,c) = (3,2,4)</math>. Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is <math>3+2+4 = boxed{9}</math>.<br />
<br />
Solution by hiker.<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2019|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Hikerhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_19&diff=1068462019 AMC 12A Problems/Problem 192019-06-23T02:27:54Z<p>Hiker: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
In <math>\triangle ABC</math> with integer side lengths,<br />
<cmath>\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \text{and} \qquad\cos C=-\frac{1}{4}.</cmath><br />
What is the least possible perimeter for <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math><br />
<br />
==Solution 1==<br />
Notice that by the Law of Sines, <math>a:b:c = \sin{A}:\sin{B}:\sin{C}</math>, so let's flip all the cosines using <math>\sin^{2}{x} + \cos^{2}{x} = 1</math> (sine is positive for <math>0^{\circ}\le x \le 180^{\circ}</math>, so we're good there).<br />
<br />
<math>\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}</math><br />
<br />
These are in the ratio <math>3:2:4</math>, so our minimal triangle has side lengths <math>2</math>, <math>3</math>, and <math>4</math>. <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
==Solution 2==<br />
<math>\angle ACB</math> is obtuse since its cosine is negative, so we let the foot of the altitude from <math>C</math> to <math>AB</math> be <math>H</math>. Let <math>AH=11x</math>, <math>AC=16x</math>, <math>BH=7y</math>, and <math>BC=8y</math>. By the Pythagorean Theorem, <math>CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}</math> and <math>CH=\sqrt{64y^2-49y^2}=y\sqrt{15}</math>. Thus, <math>y=3x</math>. The sides of the triangle are then <math>16x</math>, <math>11x+7(3x)=32x</math>, and <math>24x</math>, so for some integers <math>a,b</math>, <math>16x=a</math> and <math>24x=b</math>, where <math>a</math> and <math>b</math> are minimal. Hence, <math>\frac{a}{16}=\frac{b}{24}</math>, or <math>3a=2b</math>. Thus the smallest possible positive integers <math>a</math> and <math>b</math> that satisfy this are <math>a=2</math> and <math>b=3</math>, so <math>x=\frac{1}{8}</math>. The sides of the triangle are <math>2</math>, <math>3</math>, and <math>4</math>, so <math>\boxed{\textbf{(A) } 9}</math> is our answer.<br />
<br />
<br />
==Solution 3==<br />
<br />
Using the law of cosines, we get the following equations:<br />
<br />
<cmath>c^2=a^2+b^2+\frac{ab}{2}</cmath><br />
<cmath>b^2=a^2+c^2-\frac{7ac}{4}</cmath><br />
<cmath>a^2=b^2+c^2-\frac{11bc}{8}</cmath><br />
<br />
Substituting <math>a^2+c^2-\frac{7ac}{4}</math> for <math>b^2</math> in <math>a^2=b^2+c^2-\frac{11bc}{8}</math> and simplifying, we get the following: <br />
<br />
<cmath>14a+11b+16c</cmath>. <br />
<br />
Note that since <math>a, b, c</math> are integers, we can solve this for integers. By some trial and error, we get that <math>(a,b,c) = (3,2,4)</math>. Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is <math>3+2+4 = boxed{9}</math>.<br />
<br />
Solution by hiker.<br />
<br />
==See Also==<br />
<br />
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{{MAA Notice}}</div>Hiker