Let $ABC$ be a triangle such that $BC=AC+\frac{1}{2}AB$. Let $P$ be a point of $AB$ such that $AP=3PB$. Show that $\widehat{PAC} = 2 \widehat{CPA}.$
Problem
Source: French TST 2005 pb 5.
Tags: trigonometry, geometry solved, geometry
Igor
27.05.2005 15:52
Main idea (I'll hide it if you don't want to see it) :
On $AB$, take $I$ such that $IC=AC$. It sufficient to prove that $PI=IC$, which follows after applying once the sine law (to eliminate $IA$) and once the cosine law (to eliminate $\widehat{BAC}$).
Anto
27.05.2005 19:52
If i am not mistaken, this problem appered in MediteraneanMC 2003. Where the official solution was very ungly solution using only the rule of sine and cosine.
Igor
27.05.2005 20:14
Ok, so here is my solution: On $AB$, take $I$ such that $IC=AC$. It sufficient to prove that $PI=IC$. But $PI=PA-AI=\frac{3}{4}AB-AI$ Hence : $IC=AC \Longleftrightarrow \frac{3}{4}AB-AC=AI\Longleftrightarrow \frac{3AB}{4AC}-1=\frac{AI}{AC}$. $IC=AC \Longleftrightarrow \frac{3AB}{4AC}-1=2.\textrm{cos}(\widehat{PAC})$ by the sine law. $IC=AC \Longleftrightarrow \frac{3AB}{4AC}-1=\frac{AB^2+AC^2-BC^2}{AB.AC}$ by the cosine law. $IC=AC \Longleftrightarrow 3AB^2-4AC.AB=4AB^2+4AC^2-4BC^2$. $IC=AC \Longleftrightarrow BC^2=(AC+\frac{1}{2}AB)^2$, and we are done.
Sailor
27.05.2005 20:23
From Steward's theorem: $CP^2=b^2+\frac{3bc}{4}$. (1) But we shall prove that: $CP=2b\cos(\frac{A}{2})$, or $CP^2=4b^2\frac{p(p-a)}{bc}$, since: $p=b+\frac{3c}{4}$ ,$p-a=\frac{c}{4}$ and using (1) we get the result.
xiaozg
28.05.2005 17:32
it's Stewart's theorem.
Ghang Hwan
01.06.2005 14:23
So cool Here is my solution which use only circles. Let Q On AC with AB=AQ and opposite with C.................1 Let R ON BC with BC* CR=AC*CQ................................2 and easy to get BP*BA=BR*B C (by length condition).........3 0.5*<BAC = <AQB (by ........1) =<ARC (by .........2) =<APC (by...........3) and WE CAN SAY "Q.E.D."