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Home » High School Contests & Programs » American Mathematics Competitions
 
"Classic"? well a good geo/trig problem..
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Silverfalcon
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#1
"Classic"? well a good geo/trig problem..
AHSME 1998 #28
In triangle ABC, angle C is a right angle and CB > CA. Point D is located on \overline{BC} so that angle CAD is twice angle DAB. If AC/AD = 2/3, then CD/BD = m/n, where m and n are relatively prime positive integers. Find m + n.

\textbf{(A)}\ 10\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 22\qquad \textbf{(E)}\ 26
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PostPosted: Thu Jan 05, 2006 8:30 pm  Back to top 
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jmerry
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#2
I just clicked here out of curiosity- I got my lowest ever score (130) on the 1998 AHSME.

On the forum page, the source appears as "AHSME 1998 #..."
On the thread, the source appears as "USA AMC 12 1998, Problem ..."
On the posting form, the source appears as "AHSME 1998 #..."

That's a bit strange.

PostPosted: Thu Jan 05, 2006 8:35 pm  Back to top 
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Silverfalcon
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#3
jmerry wrote:
I just clicked here out of curiosity- I got my lowest ever score (130) on the 1998 AHSME.

On the forum page, the source appears as "AHSME 1998 #..."
On the thread, the source appears as "USA AMC 12 1998, Problem ..."
On the posting form, the source appears as "AHSME 1998 #..."

That's a bit strange.


AHSME is the correct name but database includes AHSME through AMC-12 so whole thing is called "AMC-12" in database. "USA" is included to indicate the country.
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http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1099591#1099591 (Actual proposal problems)

PostPosted: Thu Jan 05, 2006 9:02 pm  Back to top 
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Elemennop
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#4
Ah, I remember doing a really bad tan triple and double angle manipulation on this...I'll let someone with a more elegant solution post theirs.

PostPosted: Thu Jan 05, 2006 9:18 pm  Back to top 
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1234567890
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#5
I've got another solution, not sure if its "elegant", though.

Incomplete
Preliminary definitions:
AC=2x
AD=3x
BD=y
<BAD=\theta
<CAD=2\theta
AE= bisector of <CAD
Thus, <CAE=\theta and DAE=\theta
Work:
By the Angle Bisector Theorem, we can WLOG assume CE=2 and DE=3. Now we have 5^2+(2x)^2=(3x)^2 by applying the Pythagorean Theorem on triangle ACD.
\Rightarrow x=\sqrt{5}\Rightarrow AC=2\sqrt5

So by applying the Pythagorean Theorem on triangle ABC, we have (BA)^2=20+(5+y)^2=45+10y+y^2.

Now we will use Stewart's Theorem.....

Sorry to leave you hanging. But I fell asleep while posting this and need to go to bed........... Sorry.

PostPosted: Thu Jan 05, 2006 10:00 pm  Back to top 
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JesusFreak197
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#6
Answer
Assume WLOG that AC=2, AD=3. Therefore, CD=\sqrt{5}. Drawing the bisector AE of \angle CAD, then by Angle Bisector Theorem we have CE=\frac{2}{\sqrt{5}} and DE=\frac{3}{\sqrt{5}}. Therefore, AE=\frac{2\sqrt{30}}{5}, and \cos x=\frac{\sqrt{30}}{6}, \sin x=\frac{1}{\sqrt{6}}. \tan 3x=\frac{k+\sqrt{5}}{2}=\frac{\sin 3x}{\cos 3x}. \sin 3x=3\sin x-4\sin ^3x and \cos 3x=4\cos ^3x-3\cos x. Plugging in our values, we have \cos 3x=\frac{\sqrt{30}}{18} and \sin 3x=\frac{7\sqrt{6}}{18}, so \tan 3x=\frac{7\sqrt{5}}{5}=\frac{k+\sqrt{5}}{2}\Rightarrow k=\frac{9}{\sqrt{5}}=BD. Therefore, \frac{CD}{BD}=\frac{\sqrt{5}}{\frac{9}{\sqrt{5}}}=\frac{5}{9}\Rightarrow 5+9=14\Rightarrow \boxed{B}.
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PostPosted: Thu Jan 05, 2006 10:36 pm  Back to top 
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probability1.01
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#7
There is a way using no trig, but it's not that much easier to carry out (and is probably harder to find). I wonder if there is a very clever way...

PostPosted: Fri Jan 06, 2006 3:31 am  Back to top 
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#8
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Let E be the point on BC for which AE bisects \angle CAD. Through similarity, we can say that CA=2\sqrt5 and DA=3\sqrt5. Apply the pythagorean theorem to triangle ACD to get CD=5. Use the Angle Bisector Theorem to triangle ACD to get CE=2 and DE=3. For simplicity, let x=BD. Apply the pythagorean theorem to tirangle ACE to get EA=2\sqrt6, following with Angle Bisector Theorem on triangle EAB to get AB=\frac{2\sqrt6x}{3}. Now we can apply the Pythagorean Theorem once last time to triangle ABC to get:

(2\sqrt5)^2+(x+5)^2=\frac{2\sqrt6x}{3}

Solving gives x=9, making BD/DC=\frac95 and m+n=14

PostPosted: Fri Jan 06, 2006 2:35 pm  Back to top 
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Octahedr0n
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#9
What is a similarity transformation? Is it possible to do one without using matrices? (I don't get Wolfram's method: http://mathworld.wolfram.com/SimilarityTransformation.html)

Thanks in advance!
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PostPosted: Tue Nov 03, 2009 2:28 pm  Back to top 
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MathWise
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#10
Octahedr0n wrote:
What is a similarity transformation? Is it possible to do one without using matrices? (I don't get Wolfram's method: http://mathworld.wolfram.com/SimilarityTransformation.html)

Thanks in advance!


If I'm not mistaken, a similarity transformation in this case basically means "If we multiply all the lengths of the figure by a real constant (creating another similar figure), the ratios will still be the same, so we can assign one of the lengths an arbitrary value." Thus, they let AC=2\sqrt5, so then AD must be 3\sqrt5 the given ratio in the problem.

PostPosted: Tue Nov 03, 2009 3:32 pm  Back to top 
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HiDN428
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#11
I understand the solution until the point where it says now apply the Angle Bisector Theorem on triangle EAB to get AB=\frac{2\sqrt6x}{3}. Could some elaborate that step.

PostPosted: Sat Nov 07, 2009 7:34 pm  Back to top 
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