perimeter of the triangle ABC

Question

Suppose that $\triangle ABC$ has side lengths $AB = 20$ and $AC = 19$. Furthermore, let the incircle $\omega$ of $ABC$ touch segments $BC$, $CA$, $AB$ at points $D$, $E$, and $F$ respectively. Let $AD$ hit $\omega$ at a point ($P \neq D$). Suppose that $DF$ intersects the circumcircle of $CDP$ at a point ($Q \neq D$). Let line $CQ$ intersect $AB$ at point $M$. If we are given that $ \frac{AM}{BM} = \frac{5}{6} $, then the perimeter of triangle $ABC$ can be expressed as $ \frac{m}{n} $ where $m$ and $n$ are relatively prime positive integers. Find the value of $m+n$.

I have tried by applying Menelaus theorem,

Menelaus in $\triangle CBM$ with $DKA$ as a transversal $\frac{CD}{DB}\cdot \frac{AB}{AM}\cdot \frac{MK}{KC} = 1$ therefore $\frac{CD}{DB} \cdot \frac{MK}{KC} = \frac{5}{11}$. Same I tried to apply in $\triangle ABD$ , with $MKC$ as a transversal, and got therefore $\frac{BC}{DC} \cdot \frac{KD}{AK} = \frac{6}{5}$. $ Perimeter(\triangle ABC) = 39 + BC$. Can anyone help me to proceed?

Answer 1

I Have 0 reputation hence am unable to add a comment. But I wanted to note that this would be possible using barycentric coordinates, since you would know coordinates of $M$ and hence $Q$ in terms of $1$. $D$ and $F$, which depends on third side; $2$. circumcircle of $CDP$, which also depends. Then solving this one var. equation you'll be done

Answer 2

I will start with a picture of the solution:

The sides are in the picture proportional with the following numbers: $$ \begin{aligned} a &=BC=338/11\ ,\\ b &=AC=220/11=20\ ,\\ c &=AB=209/11=19\ , \end{aligned} $$ and it is for my geometrical and esthetic taste a very ugly situation. This etiquette should go directly to the author of the problem, and to all authors that even further insist to hide the answer to the problem by not asking for $a$, but for the value $m+n$ from the expression $a+b+c=m/n$ of the perimeter written as an irreducible rational fraction with positive denominator. I suppose the answer (only) is published, and the solution still remains a mystery, or should remain... Maybe this is the best method to keep young potential mathematicians away from mathematics for all times. And yes, these words above are part of the solution, it is the psychological preparation for an exam behind the problem.

I will give two solutions to the problem. Both of them are using a general picture, which allows to better distinguish lines and points in it. The first solution uses an inversion centered in $D$, since $D$ is the point common to the two "difficult" circles involved in the picture, and since the inversion is transforming them into lines, in such a manner that "other imminent objects" are also nicely transformed. The second solution uses barycentric coordinates. Feel free to compare which kind of solution should be the one to start with in an exam or in an olympiad.

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First solution: Let us draw a general picture for some specific (other) choice of the sides $b,c$ of the triangle $\Delta ABC$.

Let $r$ be the radius of the incircle in the given triangle with sides $a,b,c$. The length of the height from $A$ is denoted by $h_A$. The semi-perimeter of $\Delta ABC$ is denoted by $s:=\frac 12(a+b+c)$, its area by $S$.

We use an inversion $X\to X^*$ centered in $D$ such that the circle $\omega$ is transformed in a line through $A$ which is parallel to $BC$. Thus the point $D_1$, taken so that $DD_1$ is an $\omega$-diameter is transformed to $D_1^*$ on this parallel. So $DD_1^*=h_A$, so we have the power $k^2$ of the inversion, which is $$ k^2=DD_1\cdot DD_1^*=2r\cdot h_A=\frac 2{as}rs\cdot ah_A=\frac{4S^2}{as}\ . $$ The point $C$ is transformed in a point $C^*$ on the ray $DC$ with $DC\cdot DC^*=k^2$. The circle $(DPC)$ is transformed in a line, which is the line $P^*C^*=AC^*$. The point $Q$ is mapped into $Q^*$, a point on both $(DF)^*=DF$, and $(PDC)^*=AC^*$. So $Q^*$ is the intersection of these two lines, $$ Q^*=DF\cap AC^*\ . $$ We draw this picture:

Menelaus for $\Delta BDF$ w.r.t. the line $Q^*AC^*$ gives the location of $Q^*$ on $DF$: $$ \tag{$1$} 1 = \frac{Q^*F}{Q^*D}\cdot \frac{C^*D}{C^*B}\cdot \frac{AB}{AF}\ . $$ Menelaus for $\Delta BDF$ w.r.t. the line $MQC$ gives the location of $Q$ on $DF$: $$ \tag{$2$} 1 = \frac{QF}{QD}\cdot \frac{CD}{CB}\cdot \frac{MB}{MF}\ . $$ From $(1)$ we can extract $DQ^*$ in terms of the data of the triangle (i.e. in terms of $a,b,c;r,s,S$ etc). This gives a corresponding formula for $DQ=k^2/DQ^*$, which inserted in $(2)$ determines $MB:MF$, and finally $MB:MA$. The algebra isolating these quantities is a mess, but straightforward. I am trying to keep it compact, we have successively: $$ \begin{aligned} AE &= AF = s-a\ ,\\ BE &= BD = s-b\ ,\\ CD &= CE = s-c\ ,\\ C^*D &= k^2/CD=k^2/(s-c)\ ,\\ C^*B &= C^*D+DB=k^2/(s-c)+(s-b)\ ,\\ DF &=2\; BD\sin\frac B2=2(s-b)\sin\frac B2\ ,\\ DF^2 &=2(s-b)^2\cdot 2\sin^2\frac B2 =2(s-b)^2\cdot (1-\cos B) \\ &=2(s-b)^2\cdot \left(1-\frac{a^2 + c^2-b^2}{2ac}\right) =(s-b)^2\cdot \frac 1{ac}(-a+b+c)(a+b-c) \\ &=\frac 4{ac}(s-a)(s-b)^2(s-c) \ ,\\ 1-\frac {DF}{DQ^*} &=\frac {Q^*F}{Q^*D} = \frac{C^*B}{C^*D}\cdot \frac{AF}{AB} = \frac{k^2+(s-b)(s-c)}{k^2}\cdot \frac{s-a}{c} \ , \\ \frac {DF}{DQ^*} &=1-\frac{k^2+(s-b)(s-c)}{k^2}\cdot \frac{s-a}{c} \ ,\\[3mm] DQ&=\frac {k^2}{DQ^*} = \frac{k^2}{DF}\left(1-\frac{k^2+(s-b)(s-c)}{k^2}\cdot \frac{s-a}{c}\right) \\ &= \frac 1{c\; DF}\Big(\ k^2c-k^2(s-a)-(s-a)(s-b)(s-c)\ \Big) \\ &= \frac 1{c\; DF}\left(\ k^2(s-b)-\frac{S^2}s\ \right) = \frac 1{c\; DF}\left(\ \frac{4S^2(s-b)}{as}-\frac{S^2a}{as}\ \right) \\ &= \frac {S^2(a-2b+2c)}{ac\; s\; DF} \ ,\\[3mm] \frac{QF}{QD} &= \frac{DF-QD}{QD} = \frac{DF}{DQ}-1 = \frac{ac\; s\; DF^2}{S^2(a-2b+2c)} -1\\ &= \frac{4s\; (s-a)(s-b)^2(s-c)}{s(s-a)(s-b)(s-c)\; (a-2b+2c)} -1 \\ &=\frac{4(s-b)}{(a-2b+2c)}-1 =\frac{2a-2b+2c}{a-2b+2c}-1 =\frac a{a-2b+2c} \ , \\[3mm] \frac{s-b}{MB}-1 &=\frac{FB}{MB}-1 =\frac{FB-MB}{MB} =\frac{FM}{MB} \\ &=\frac{QF}{QD}\cdot\frac{CD}{CB} = \frac a{a-2b+2c}\cdot \frac{s-c}a = \frac {a+b-c}{2a-4b+4c}\ , \\ \frac{s-b}{MB} &= \frac {a+b-c}{2a-4b+4c}+1 = \frac {3a-3b+3c}{2a-4b+4c} = \frac {3(s-b)}{a-2b+2c} \ , \\ MB &=\frac 13(a-2b+2c)\ ,\\ AM &= AB-MB =c-\frac 13(a-2b+2c)=\frac 13(-a+2b+c)\ ,\\ \frac{AM}{MB} &=\frac{-a+2b+c}{a-2b+2c}\ . \end{aligned} $$ It remains to plug in the known values for $b,c$, and solve for $a$ in the equation obtained from $AM:MB=5:6$.

$\square$

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Second solution: We are using barycentric coordinates. A point $W$ has the (normed) coordinates $(x,y,z)$ if $W=xA+yB+zC$, and there is also the norming condition $x+y+z=1$. The relation $W=xA+yB+zC$ has sense either by using vectors, so that for some / any point $O$ in the plane we have vectorially $OW=xOA+yOB+zOC$, or by using affixes in the complex plane.

Homogeneous coordinates are used often instead, we do not insist to have the norming condition, but only $x+y+z\ne 0$, and the point $[x:y:z]$ is the point with normed coordinates $(\ x/(x+y+z),\ y/(x+y+z),\ z/(x+y+z)\ )$. We compute now equations for the involved objects: $$ \begin{aligned} A &=(1,0,0)\ ,\\ B &=(0,1,0)\ ,\\ C &=(0,0,1)\ ,\\ D &=[0:s-c:s-b]\ ,\\ E &=[s-c:0:s-a]\ ,\\ F &=[s-b:s-a:0]\ ,\\ \odot(DEF)\ &:\ 0=-a^2yz-b^2zx-c^2xy+(x+y+z)\Big((s-a)^2x+(s-b)^2y+(s-c^2)z\Big)\ ,\\ AD\ &:\ y(s-b)=z(s-c)\ ,\\ P & =\left[\ \frac 4{b+c-a}\ :\ \frac 1{a-b+c}\ :\ \frac 1{a+b-c}\ \right] =\left[\ \frac 4{s-a}\ :\ \frac 1{s-b}\ :\ \frac 1{s-c}\ \right] \ , \\ \odot(CDP)\ &:\ 0=-a^2yz-b^2zx-c^2xy+(x+y+z)\Big(ux+vy+wz\Big)\ ,\\ &\qquad\text{ and we determine $u,v,w$.} \\ &\qquad\text{ Since $C=(0,0,1)$ is on $\odot(CDP)$ we have $w=0$.} \\ &\qquad\text{ Since $D\in\odot(CDP)$, plugging in its components gives} \\ &\qquad\text{ $0=-a^2(s-b)(s-c)-0-0+a(0+v(s-c)+0)$, so $v=a(s-b)$.} \\ &\qquad\text{ Plugging in $P$ determines finally $u=(s-a)^2$. So:} \\ \odot(CDP)\ &:\ 0=-a^2yz-b^2zx-c^2xy+(x+y+z)\Big((s-a)^2x+a(s-b)y\Big)\ , \\ DF\ &:\ 0= \begin{vmatrix} x & y & z\\ 0 & s-c & s-b\\ s-b & s-a & 0 \end{vmatrix}\ ,\text{ i.e.}\\ DF\ &:\ 0= -x(s-a) + y(s-b) + z(s-c)\ , \\ Q &=[\ a-2b+2c\ :\ -a +2b+c\ :\ c\ ]\ , \\ M &=[\ a-2b+2c\ :\ -a +2b+c\ :\ 0\ ]\ . \end{aligned} $$ The given relation $AM:BM = 5:6$ reads $M=[6:5:0]$. So we obtain the needed missing equation, which is linear: $$ 5(a-2b+2c)=6(-a+2b+c)\ . $$ Solving we obtain the solution initially claimed.

$\square$