Prove that CAP = 2 CPA

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Prove that CAP = 2 CPA

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#1 h Oct 31, 2010, 2:18 PM • 2 Y

Y by Adventure10, Mango247

In a triangle $ABC$ with $BC = CA + \frac 12 AB$ , point $P$ is given on side $AB$ such that $BP : PA = 1 : 3$ . Prove that $\angle CAP = 2 \angle CPA.$

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yetti

#2 h Nov 1, 2010, 11:35 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Lemma: $\{X \in (0, \pi) | f(X) = \frac{\sin (\Phi-X)}{\sin X} \}$ is decreasing for any $\Phi \in (0,\pi),$ because its derivative $f'(X) = -\frac{\sin \Phi}{\sin^2 X} < 0.$

$CA + \frac{_1}{^2} AB = BC \Longleftrightarrow$

$\frac{3}{4} \cdot \frac{AB}{CA} = \frac{CA^2 + CA \cdot AB + AB^2 - BC^2}{CA \cdot AB} \Longleftrightarrow$

$\frac{\sin \angle ACP}{\sin \angle CPA} = \frac{AP}{CA} = 1 + 2\cos A = 3 - 4 \sin^2 \frac{A}{2} = \frac{\sin \frac{3A}{2}}{\sin \frac{A}{2}} = \frac{\sin (\pi - \frac{3A}{2})}{\sin \frac{A}{2}}.$

In addition, $\angle ACP + \angle CPA = \pi - A = \pi - \frac{3A}{2} + \frac{A}{2}.$ By the lemma, $\angle CPA = \frac{A}{2}.$

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jgnr

#3 h Nov 2, 2010, 1:52 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Take $M\in(BC)$ such that $CA=CM$ . Let $CA=CM=b$ , $BP=\frac13AP=\frac12BM=a$ . By Stewart's theorem, $CP^2=CB^2\cdot\frac34+CA^2\cdot\frac14-AP\cdot PB=(b+2a)^2\cdot\frac34+b^2\cdot\frac14-3a^2=b(b+3a)=CA(CA+AP).$ It is well-known that $\angle CAP=2\angle CPA\Leftrightarrow CP^2=CA(CA+AP)$ , hence our proof is complete.

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#4 h Nov 7, 2010, 3:50 AM • 2 Y

Y by Adventure10, Mango247

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=371578

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#5 h Nov 27, 2011, 5:13 AM • 2 Y

Y by Adventure10, Mango247

it hsas the following nice analytical form:
let C be a point on the left branch of a hyperbola with focuses A,B,P is its right vertex.
prove that $\angle CAP=2\angle CPA$ .

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silvergrasshopper

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silvergrasshopper

#6 h Jun 18, 2012, 4:47 AM • 2 Y

Y by Adventure10, Mango247

Johan Gunardi wrote:

$\angle CAP=2\angle CPA\Leftrightarrow CP^2=CA(CA+AP)$

Where does this come from?

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Sayan

#7 h Jun 18, 2012, 4:54 AM • 2 Y

Y by Adventure10, Mango247

silvergrasshopper wrote:

Johan Gunardi wrote:

$\angle CAP=2\angle CPA\Leftrightarrow CP^2=CA(CA+AP)$

Where does this come from?

see here
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=54513&

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drEdrE

#8 h Jun 19, 2012, 6:20 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Or : calculate PC using Stewart's Theorem, then apply sinus theorem in APC. The problem becomes equivalent with $2 \cos ( \frac{A}{2})=\frac{PC}{AC}$ . Square this relation, and use one of Neper Formulas for cos(A/2) : $\cos ( \frac{A}{2})= \sqrt{\frac{p(p-a)}{bc}}$ .

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#9 h Jun 19, 2012, 8:47 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

You may see here as well.

Best regards,
sunken rock

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