Geometry Challenge: Parallelogram in $\triangle$ $ABC$ with circumscribed Circle

Question

I hope this message finds you in high spirits. I am writing to seek your expertise in solving a captivating geometry problem that I recently encountered in a competitive exam. Despite my best analytical efforts, the solution has proven elusive. Your expertise in this matter would be greatly appreciated, and I look forward to the insights that will undoubtedly enhance my understanding of this geometric challenge.

$ \textbf{Problem Description:} $

Consider $\triangle$$ ABC $, where $ M $ is the midpoint of side $ AC $, and $ O $ is the center of a small circle with the diameter $BM$. Let $ BPTQ $ be a parallelogram formed by joining points $ B $, $ P $, $ T $, and $ Q $. This question explores a geometric relationship within this configuration. Given the setup in $\triangle$$ ABC $ and parallelogram $ BPTQ $, we want to find an expression for $ \left(\frac{BT}{BM}\right)^8 $.

My Efforts:

I've tried various analytical approaches, but my results have been inconsistent. A more systematic strategy involving geometric properties, algebraic manipulation, or other mathematical tools seems necessary.

I'm sincerely thankful for your time and insights. Feel free to ask if you need further details.

Appreciate your help greatly!

Answer 1

Incomplete solution. SEE EDIT and COMPLETE SOLUTION below

Your phrasing is really confusing. If you just draw from $P$ and $Q$ the lines parallel to $BC$ and $AB$ respectively, they will meet at a point $T$, forming a parallelogram. There are many configurations in which $T$ lies on $(ABC)$.

From your drawing and video, though, it appears that $T$ must lie on the perpendicular bisector of $AC$, which forces the arcs $\overset{\huge\frown}{AT}$ and $\overset{\huge\frown}{TC}$ to be equal. In that case $BT$ bisects $\angle ABC$, and therefore $BPTQ$ is in fact a rhombus, and the solution becomes unique.

In conclusion what we can show is the following.

Let $ABC$ be a triangle with median $BM$. Let $P$ and $Q$ be the intersections between the circle with diameter $BM$ and lines $AB$ and $BC$ respectively. Let also $T$ be the intersection between the perpendicular bisector of $AC$ and the circumcircle of $ABC$. Show that if $BPTQ$ is a parallelogram, then $BT/BM = \sqrt 2$.

Let $O$ be the center of the rhombus and $O_1\in BM$ be the center of $(BPQ)$.

1. As stated earlier $\overset{\huge\frown}{AT} \cong \overset{\huge\frown}{TC}$ implies $\angle ABT \cong\angle TBC$.
2. Then by ASA criterion $BTP \cong BTQ$. Hence $BPTQ$ is a rhombus.
3. By SSS criterion we have $BPO_1 \cong BQO_1$. Therefore $\angle ABM \cong \angle MBC$.
4. Comparing 1. and 2. we obtain that $B$, $T$, and $M$ are aligned.
5. Since $BM$ is both median and internal bisector, $ABC$ is an isosceles triangle, and $BT$ perpendicularly bisects $AC$.
6. From 5. we derive that $BT$ is a diameter of $(ABC)$ and hence $O$ is the center of $(ABC)$.
7. WLOG let the smaller circle diameter be $\overline{BM} = 2$. If $\overline{BO} = x$. Then, similarity $BPO\sim MPO$ implies $$\overline{PO} = \sqrt{x(2-x)}.$$
8. Similarity $BPO\sim BAM$ and 7. give $$\overline{AM} = 2\sqrt{\frac{2-x}{x}}.$$
9. Now observe that $AO \cong BO$, so that Pythagorean Theorem on $AOM$ leads to the equation $$(2-x)^2 + \frac{4(2-x)}{x} = x^2,$$ whose only valid solution is $$x = \sqrt 2.$$ This yields $$\boxed{\frac{\overline{BT}}{\overline{BM}} = \sqrt{2}}.$$

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EDIT

Thanks to David K's comment, I noticed that all the configurations (not only the one in which $T$ bisects arc $AC$) lead to the same ratio $\sqrt 2$. I will try to work an a more complete solution. I leave for now the answer, though incomplete: it may be useful for others to work out the entire problem.

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SOLUTION TO THE GENERAL CASE

Probably a bit of an overkill, but I could not find any better approach.

Suppose that point $T$ is determined as the fourth vertex of a parallelogram $BPTQ$. We want to show that if $T \in (ABC)$ then $\overline{BT}/\overline{BM} = \sqrt 2$.

Let us fix a coordinate system with the origin in $B$, and let, WLOG, $C(2,0)$. Suppose $A$ lies on the line with equation $y=mx$, and precisely let $A(2t,2mt).$

Then the perpendicular bisector of $AB$ has equation $y-mt=-\frac1m(x-t)$. From here, we can easily find the circumcenter of $(ABC)$, that is $$O'\left(1,\frac{-1+t+m^2t}m\right),$$ and thus the equation of $(ABC)$ $$x^2+y^2-2x +2y\frac{1-t-m^2t}{m}=0.\tag{1}\label{1}$$ We also have $M(1+t,mt)$ and therefore $$\overline{BM}^2 = (1+t)^2+ m^2t^2.\tag{2}\label{2}$$ Point $Q$ is the projection of $M$ on $BC$. So we get $Q(1+t,0)$. Point $P$ is the projection of $M$ on $AB$. Intersection between line $y=mx$ and line $y-mt = -\frac{1}{m}(x-t-t)$ yields $$P\left(\frac{m^2t+1+t}{m^2+1},m\frac{m^2t+1+t}{m^2+1}\right).$$ It is now straightforward to intersect lines $y=m\frac{m^2t+1+t}{m^2+1}$ and $y=m(x-1-t)$ to get $$T\left(\frac{2m^2t+m^2+2+2t}{m^2+1},m\frac{m^2t+1+t}{m^2+1}\right).$$ Since $T$ lies on $(ABC)$, its coordinates satisfy \eqref{1}, which means, in particular, that $$\overline{BT}^2 = x_T^2+y_T^2 = 2x_T-2y_T\frac{1-t-m^2t}{m}.\tag{3}\label{3}$$ Replacing in \eqref{3} the coordinates of $T$ we just found we derive \begin{eqnarray} \overline{BT}^2&=&2\cdot \frac{2m^2t+m^2+2+2t-(1+m^2t+t)(1-m^2t-t)}{m^2+1}\\ &=& 2 \cdot \frac{2m^2t+m^2+2+2t-[1-(m^2t+t)^2]}{m^2+1}=\\ &=&2 \cdot \frac{2t(m^2+1)+(m^2+1)+t^2(m^2+1)^2}{m^2+1}=\\ &=& 2 \cdot (2t+1+t^2+m^2t^2)=\\ &\stackrel{\eqref{2}}{=}& 2\overline{BM}^2. \end{eqnarray}