A geometry problem about triangle angles and perimeter Consider $\Delta ABC$ with three acute  angles, we draw its altitudes and make $\Delta MNP$ triangle
if $\frac{PN}{KN}=\frac{3}{2}$ and $\frac{\sin{\alpha}}{\cos{\frac{\alpha}{2}}}+\frac{\sin{\theta}}{\cos{\frac{\theta}{2}}}+\frac{\sin{\gamma}}{\cos{\frac{\gamma}{2}}}=\frac{288}{100}$ then calculate $\frac{MN}{AB+BC+CA}$
Note that $\alpha,\theta,\gamma$ are angles of $\Delta MNP$ and $K$ is the point of concurrency of $MN$ and $CP$

I think it is a famous geomtry problem, I can't remember where I saw this first time but I think it was a famous question...
I thought on this problem a lot but I have no idea to solve that, except that the fraction $\frac{288}{100}$ is $2*\frac{144}{100}$ and I think I should use of this... Maybe I should radical this fraction.
Am I right? 
 A: I will attempt to solve this problem with as little trigonometry as possible. 
The value of $\frac{144}{100}=(1.2)^2$ is actually a red herring. First we note by $a$, $b$, $c$, $\angle{A}$, $\angle{B}$, $\angle{C}$, $S$, $R$ and $r$ the sides, angles, area, circumradius and inradius of $ABC$. Note that: $$\frac{\sin \alpha}{\cos \frac{\alpha}{2}}=\frac{2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}=2\sin\frac{\alpha}{2}=2\sin\frac{\angle{NMP}}{2}=2\sin\angle{AMP}=2\sin\angle{AMP}=2\sin\angle{NBA}=2\sin(90^{\circ}-\angle{AMP})=2\cos\angle{BAC}=2\cos A$$
So we have that $\cos A+\cos B+\cos C=\frac{144}{100}$. Now we will prove that in any triangle we have $\cos A+\cos B+\cos C = 1+\frac{r}{R}$. It can be proven in many ways but one of the nicer ones is this: 
Consider the midpoints $D$, $E$, $F$ of $BC$, $CA$, $AB$ respectively which are also the projections of point $O$ - the circumcentre of $ABC$ onto its sides. Denoting by $x$, $y$ and $z$ the lengts of $OD$, $OE$, $OF$ and applying Ptolemy theorem to the cyclic quadrilateral $AEOF$ we obtain:
$$AE \cdot OF + AF \cdot OE = AO \cdot EF$$
$$\frac{b}{2} \cdot z + \frac{c}{2} \cdot y = R \cdot \frac{a}{2}$$
$$bz+cy=aR$$
Writing analogous equations and adding them up we get:
$$x(b+c)+y(c+a)+z(a+b)=R(a+b+c)$$
Since $ax$ is twice the area of $BOC$ and similarly for $by$ and $cz$, $ax+by+cz=2S$ and so:
$$(x+y+z)(a+b+c)=x(b+c)+y(c+a)+z(a+b)+(ax+by+cz)=R(a+b+c)+2S$$
dividing by $(a+b+c)$ and using the fact that $2P=r(a+b+c)$ we get:
$$x+y+z=r+R$$
It's a nice result, but how does it connect to our sum of cosines? Just notice that $\angle{DOB}=\frac{1}{2}\angle{BOC}=A$ so in triangle $BOD$ we have $\cos A=\cos \angle{DOB}=\frac{DO}{OB}=\frac{x}{R}$. Writing analogous equations we obtain:
$$\cos A + \cos B + \cos C = \frac{x}{R}+\frac{y}{R}+\frac{z}{R}=\frac{x+y+z}{R}=\frac{R+r}{R}=1+\frac{r}{R}$$
OK, so far we have $\frac{144}{100}=\cos A+\cos B+\cos C=1+\frac{r}{R}$ so $\frac{r}{R}=0.44$. 
Now we will derive the formula for the perimeter of triangle $MNP$. To do this note that reflecting $M$ across $AB$ and $AC$ results in points $Y$ and $Z$ which lie on $PN$. Moreover we have:
$$MN+NP+PM=ZN+NP+PY=YZ$$
So this perimeter is equal to the length of $YZ$. Its half is therefore equal to the length of $Y'Z'$ where $Y'$ and $Z'$ are midpoints of $MY$ and $MZ$ which are also projections of $M$ onto $AB$ and $AC$. Now if we define $A'$ as the antipode of $A$ on the circumcircle of $ABC$ we can say that the quadrilaterals $AY'MZ'$ and $ACA'B$ are (inversely) similar. This in turn yields that the ratios of their diagonals are equal i.e.:
$$\frac{Y'Z'}{AM}=\frac{BC}{AA'}=\frac{a}{2R}$$
Since $a \cdot AM = 2S$ we have:
$$MN+NP+PM=YZ=2Y'Z'=\frac{2AM \cdot a}{2R}=\frac{4S}{2R}=\frac{2S}{R}=\frac{(a+b+c)r}{R}$$
That means that the ratio of the perimeters of $MNP$ and $ABC$ is $\frac{r}{R}=0.44$. 
Now let's tackle our main problem - by the angle bisector theorem we have:
$$\frac{3}{2}=\frac{PN}{KN}=\frac{PM}{KM}=\frac{PN+PM}{KN+KM}=\frac{PN+PM}{MN}=\frac{PN+PM+MN}{MN}-1$$
Where in the middle we used the fact that if $\frac{a}{b}=\frac{c}{d}$ then their common value is also equal to $\frac{a+c}{b+d}$. So:
$$\frac{MN}{PN+PM+MN}=\frac{2}{5}=0.4$$
And finally:
$$\frac{MN}{AB+BC+CA}=\frac{MN}{a+b+c}=\frac{MN}{PN+PM+MN} \cdot \frac{PN+PM+MN}{a+b+c}=0.4 \cdot 0.44=0.176$$ 
A: I will use some idetities or properties, you can just google them if you don't know them.
$$1.PN=a\cos A,NM=c\cos C,MP=b\cos B$$
$$2.a\cos A+b\cos B+c\cos C=2a\sin B\sin C$$
$$3.\cos A+\cos B+\cos C=1+4\sin {A\over 2}\sin {B\over 2}\sin {C\over 2}, \sin A+ \sin B+\sin C=4\cos {A\over 2}\cos {B\over 2}\cos {C\over 2} $$
Also, some basic properties like Law Of Sines$(4)$ and the Angle Bisector Theorem$(5)$ is used.
Now let's start.
Denote the perimeter of $\triangle NMP$ as $l$, and $AB+BC+CA=s$
First, according to $(5)$,$${PN \over NK}={PM \over MK}={3/2}$$
we know that $${MN\over l}={2 \over 5}$$
Second, notice that $\alpha, \theta,\gamma$ are only a permutation of $A+B-C ,A+C-B ,B+C-A $.
And $${A+B-C\over 2}={\pi \over 2}-C$$
We have $${72 \over 25}=2\sin {\alpha \over 2}+2\sin {\theta \over 2}+2\sin {\gamma \over 2}=2(\cos A+\cos B+\cos C)$$
So $$\cos A+\cos B+\cos C={36 \over 25}$$
Third, we calculate $l \over s$.
From $(1),(2)$ we know that $$l=a\cos A+b\cos B+c\cos C=2a\sin B\sin C$$
So$${l \over s}={2a \sin B\sin C \over a+b+c}={2\sin A \sin B\sin C \over \sin A+ \sin B+\sin C}={16\sin {A\over 2}\sin {B\over 2}\sin {C\over 2}\cos {A\over 2}\cos {B\over 2}\cos {C\over 2}\over 4\cos {A\over 2}\cos {B\over 2}\cos {C\over 2}}
=4\sin {A\over 2}\sin {B\over 2}\sin {C\over 2}=\cos A+\cos B+\cos C-1={11 \over 25}$$
Notice that $(3)$ is used several times in the last few steps.
Now finally, we have$${MN \over l}={2 \over 5}, {l \over s}={11 \over 25}$$
So $${MN \over s}={22 \over 125}$$,and we are done.
