Which parameters can we choose in order to solve this triangle construction issue

Question

This is follow-on of a question asked yesterday, with real work shown under the form of sketches but not understandable. Visibly, the asker isn't used to formulate mathematics with sentences (his/her first question on this site). On this ground, this question has been closed.

The question is interesting, in particular necessitating a non-evident investigation about the choice of parameters describing the different configurations (not being sure that a single solution exists ...) and as I finally found a way to solve it (see answer), I chose to reproduce it and give an answer, wishing exchanges with other people having a different approach.

Here is the text of the exercise :

Let $ABC$ be an acute-angle triangle ; let $D$ be the foot of the altitude issued from vertex $A$, and let $M$ be the midpoint of side $[AC]$. A point $P$ is taken on median $[BM]$ in such a way that $∠PAM = ∠MBA$. If $∠BAP=41°$ and $∠PDB=115°$ (expressed in degrees), what is the value of angle $∠BAC$ ?

Answer 1

Very nice exercise of similarities plus angle chasing. This is a formidable example, I believe, of how much a "synthetic" approach can simplify the matter with respect to an "analytic" one. Note that you can play around with the angles' measures as you like: only the last step, in fact, involves them!

1. By AAA criterion $ABM \sim APM$. Implying that $PM:AM=AM:BM$.
2. Since $AM \cong MC$ we have $PM:MC = MC:BM$, involving that $BMC \sim PMC$ by SAS criterion. In particular \begin{equation}\angle MCP \cong \angle MBC,\tag{1}\label{1}\end{equation} and \begin{equation}\angle MPC \cong \angle MCB.\tag{2}\label{2}\end{equation}
3. Since $ADM$ is right-angled, we also have $MD \cong MC$, yielding $PM:MD=MD:BM$, that is $BMD \sim PMD$ (again SAS criterion). Hence \begin{equation}\angle BDM \cong \angle DPM.\tag{3}\label{3}\end{equation}
4. Since $DMC$ is isosceles and because of \eqref{2}, we get the congruence $$\angle MDC \cong \angle MPC,$$and consequently the one involving the supplementary angles, i.e. \begin{equation}\angle BDM \cong \angle BPC.\tag{4}\label{4}\end{equation}
5. By \eqref{3} and \eqref{4}, $\angle BPC \cong \angle DPM$, and, by subtracting $\angle DPC$ to both, we get \begin{equation}\angle BPD \cong \angle MPC,\tag{5}\label{5}\end{equation} as shown in the Figure above.
6. Conditions \eqref{1} and \eqref{5} yield $BPD \sim MPC$ (AAA criterion), so that $$\measuredangle PMC = 115^\circ.$$
7. Using this last information and Exterior Angle Theorem on $ABM$ we get $$41^\circ + 2x = 115^\circ,$$ that is $$\boxed{x=37^\circ}.$$

Answer 2

Fig. 1.

Here is a solution mainly based on analytical geometry.

Let us take the origin in $D$ and axes defined by $\vec{DC}$ and $\vec{DA}$. Which units do we take on these axes ?

As this issue is invariant by scaling (any triangle which is a solution generates an infinite number of solutions by enlargement or shrinking), and because the triangle is with acute angles, we can assume WLOG that $D(-1,0)$.

Therefore, if we denote by $(p,h)$ the coordinates of $M$, the coordinates of $C$ and $A$ will be $C(2p,0)$ and $A(0,2h)$ resp. (using an elementary property of right triangle $ADC$).

It remains to choose a way to describe the position of $P$ on line segment $BP$ ; it will be done in a barycentric way thus :

$$P=(1-m)B+mM \ \text{where} \ 0 \le m \le 1\tag{1}$$

(Intuition : the closer $m$ is to $1$, the closer $P$ is to $M$).

We are going to show that variables $p,h$ and $m$ are sufficient to describe the issue.

Let us call momentarily $(x,y)$ the coordinates of $P$.

The definition of $P$ (given in (1)) together with a slope constraint (the slope of $DP$ is clearly $t:=\tan(65°)$) gives the following 2 ways to express the coordinates of $M$

$$(x,y)=(\frac{mh}{t},mh)=(-(1-m)+mp,mh)$$

Comparison of abscissas yields the first constraint connecting variables $m,p$ and $h$ :

$$Eq. 1: \ \ \frac{mh}{t}=-(1-m)+mp$$

Now, let us "hunt" for 2 other equations :

The first one deals with the fact that we can express in two ways the dot product between $\vec{AB}$ and $\vec{AP}$ :

$$\vec{AB}.\vec{AP}=\underbrace{\cos(41°)}_{c} \|\vec{AB}\|\|\vec{AP}\|$$

Squaring this relationship gives :

$$Eq. 2: \ \ ((m/t)+2h(m-2))^2 = c^2((m/t)^2+(m-2)^2)(1+4h^2)$$

(I can provide details...)

Constraints on angle $115°$, resp angle $41°$ have been taken into account in Eq. 1, resp. Eq. 2. It remains to encapsulate into a certain equation the fact that $∠PAM=∠MBA=x$. This should be done without introducing this unknown angular common value $x$ as a new parameter. After different attempts, I have done it by expressing a certain ratio of triangles' area in two ways :

$$\frac{area(APM)}{area(ABM)}=\frac{PM}{BM},$$

(indeed these triangles share the same altitude issued from $A$). Therefore :

$$\frac{\tfrac12 \sin(x).AM.AP}{\tfrac12 \sin(x).BA.BM}=1-m\tag{3}$$

In this way, the unknown quantity $\sin(x)$ is cancelled ; it remains to square this relationship and express it with coordinates, giving :

$$Eq. 3: \ \ \frac{h^2((m/t)^2+(m-2)^2)(1+h^2)}{(k^2+4h^2)((1+k)^2+h^2)}=(1-m)^2$$

I have implemented these equations into a "Sage" program (see below).

var('h p m k');
t=n(tan(65*pi/180));
c=n(cos(41*pi/180));
eq1 = m*h==t*(m*p-(1-m));
#eq1 = m==t/(t+t*p-h);
eq2 = (1+4*h^2)*((1+p)^2+h^2)*(1-m)^2==((m*h/t)^2+(m*h-2*h)^2)*(p^2+h^2);
eq3 = ((m/t)+2*h*(m-2))^2 == c^2*((m/t)^2+(m-2)^2)*(1+4*h^2);
eq=[eq1,eq2,eq3];
sol=solve(eq,h,p,m);
[show(sol[k]) for k in range(len(sol))]

which has given a lot of solutions (indeed, by squaring, as we have done for the second and third equation, we introduce spurious solutions). Only one triple $(p,h,m)$ respects the domains $$p>0,h>0,0 \le m \le 1.$$ It is

$$p \approx 1.0275315, h \approx 0.89727272, m \approx 0.62145522.$$

Remarks :

1. The first equation (Eq. 1) allows to express $m=t/(tp+t-h)$ as a function of $p$ and $h$ (it can be seen as a comment in the "Sage" program), $t$ being a constant. In this way, one could replace all occurences of $m$ in Eq. 2 and Eq. 3 by the fraction above , giving a description by the two unknowns $p$ and $h$ only. Geometrical interpretation as displayed on Fig. 2:

Fig. 2 : Blue curve corresponds to Eq. 2, Red curve to Eq. 3. The looked for solution is point $(p,h)\approx (1.03,0.90)$ situated in the first quadrant.

point $M$ is situated at the intersection of 2 curves in a $(p,h)$ plane, which is not different from the plane in which this question is asked...

2. Having all the coordinates of all the points, one can compute (using the sines' relationship in triangle $ABM$): $x=∠MBA=37°$, showing that $∠BAC=41°+37°=78°$.

3. There must exist a solution using barycentric coordinates.

Answer 3

In fact, I have discovered that a single parameter can be taken, and this parameter is directly in connection with the looked-for angle.

Let us have a look at the following picture :

You can see that the initial picture has been mirrored, $B$ is taken as the origin, line segment $BA$ is considered as reference length $1$ (otherwise said, $A$ has abscissa $1$), with polar reference axis directed by $\vec{BA}$. All this can be done WLOG (because the scaling of the figure is arbitrary).

Two main ideas :

• work on the locus of $M$ (blue curve), then the locii of $C$, $D$ and $P$ (red, green, yellow resp.).

• describe these loci starting with the polar equation of $M$ as a function of $\theta$ (the angle named $x$ in the question).

One can establish that the polar equation of $M$ is

$$r=\underbrace{\frac{\sin(\theta+a)}{\sin(2\theta+a)}}_{\text{length of BM}} \ \ \iff \ \ z_M=\frac{\sin(\theta+a)}{\sin(2\theta+a)}e^{i \theta}$$

where $a=41°$ (or its equivalent in radians...).

($r$ is obtained by angle chasing in triangles $APB$ and $BPM$ followed by an application of sine law in triangle $AMB$).

Having this generic description of $z_M$, one can deduce the complex numbers :

• $z_C=2z_M-1$ associated with $C$.

• $z_P=k z_M$ associated with $P$ with $k=\frac{1}{1+|1-z_M|}\frac{\sin(\theta)}{\sin(a)}$

• $z_D=\frac{\cos(arg(z_C))}{|z_C|}z_C$ associated with $D$.

(of course, $D$ describes an arc of the circle centered in $(0,1/2)$ with radius $1/2$).

As we have used complex numbers, the idea is to choose the value of $\theta_0$ in such a way that $arg(\frac{z_C-z_D}{z_P-z_D})=65°$. We find by trial and error that $\theta = 37°$ (an exact integer value).

Remark : I use the expression "trial and error" ; in fact, I am almost sure that there exist a classical numerical method giving directly the solution.

I don't enter into the details of the computation of the expressions of $z_M,z_C,z_P,z_D$: my objective was/is to show how one can tackle this kind of issue, with an interesting parametrization ; I consider my objective as reached because this second solution uses, right from the beginning, a single - and natural - parametrization.

Matlab program having generated the figure :

clear all;close all;axis equal;hold on;
axis([0,1,0,1.5]);
LW='linewidth';ep=0.00001;O=ep+i*ep;
cf=pi/180; % conv. degrees <-> radians
a=41*cf;
t0=37*cf; % the good value : 37°
t=0:0.01:(pi-a)/2;
M=@(th)((sin(th+a)./sin(2*th+a)).*exp(i*th))
C=@(th)(2*M(th)-1);
P=@(th)(M(th)./(1+abs(1-M(th)).*sin(th)/sin(a)));
D=@(th)(cos(angle(C(th))).*C(th)./abs(C(th)));
traj=[O,O+1,D(t0),O+1,C(t0),O,P(t0),M(t0)];
set(gcf,'defaulttextfontsize',14)
Zx=zeros(1,8);Zx(2)=0.12;Zx(8)=0.12;Zx(7)=0.06;
text(real(traj)-0.08+Zx,imag(traj)+0.05,{'B','A','D','','C','','P','M'});
plot(traj,'k',LW,5);
plot(0.5+0.5*exp(i*(0:0.01:pi)),'k',LW,1);
text(0.14,0.07,'O');plot([0.15,0.20],[0.07,0.07],'k',LW,4)
plot(D(t),'y',LW,5);plot(D(t0),'ok');
plot(P(t),'g',LW,5);plot(P(t0),'ok');
plot(M(t),'b',LW,5);plot(M(t0),'ob');
plot(C(t),'r',LW,5);plot(C(t0),'or');
angle((C(t0)-D(t0))/(P(t0)-D(t0)))/cf, % value 65°
% (recall 65° = 180° -115°)