# 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?

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: $$\frac{MN}{AB+BC+CA}=\frac{MN}{a+b+c}=\frac{MN}{PN+PM+MN} \cdot \frac{PN+PM+MN}{a+b+c}=0.4 \cdot 0.44=0.176$$

• Dear friend: When you say that $\angle{\frac{NMP}{2}}= \angle AMP$, you are assuming that the height $AM$ of the triangle $\triangle ABC$ is the bisector of the angle $\angle NMP$. I am afraid that this is not true and that here it is just an optical illusion of the drawing presented by the O. P. Am I wrong? – Piquito May 8 at 0:59
• $AM$ is indeed a bisector of $\angle{NMP}$. To see that notice that quadrilaterals $HPBM$ and $HNCM$ are cyclic ($H$ denotes the orthocentre of $ABC$). From this follows:$$\angle{AMP}=\angle{HMP}=\angle{HBP}=\angle{HBA}=90^{\circ}-A=\angle{HCA}=\angle{HCN}=\angle{HMN}=\angle{AMN}$$ – Bartek May 8 at 1:30
• See however the attached figure. Regards. – Piquito May 8 at 2:05
• Dear Piquito, I'm afraid that you have made a typo - it should have been $\frac{1.906}{8.576-7}$ and both values are equal to $1.208980044$. – Bartek May 8 at 2:21
• Maybe and if this has been the case I am happy for you. – Piquito May 8 at 2:28

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.