1
$\begingroup$

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 $CP$

enter image description here

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?

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Piquito May 8 at 0:59
  • $\begingroup$ $AM$ 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:$$\angle{AMP}=\angle{HMP}=\angle{HBP}=\angle{HBA}=90^{\circ}-A=\angle{HCA}=\angle{HCN}=\angle{HMN}=\angle{AMN}$$ $\endgroup$ – Bartek May 8 at 1:30
  • $\begingroup$ See however the attached figure. Regards. $\endgroup$ – Piquito May 8 at 2:05
  • $\begingroup$ 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 $1.208980044$. $\endgroup$ – Bartek May 8 at 2:21
  • $\begingroup$ Maybe and if this has been the case I am happy for you. $\endgroup$ – Piquito May 8 at 2:28
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.