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Let $ABC$ be a right triangle acutangle and let $\overline{ AD}$, with D in $\overline{BC}$, be a height relative to point $A$. Let $Γ_1$ and $Γ_2$ as circumferences circumscribed to triangles $ABD$ and $ACD$, respectively. The circumference $Γ_1$ crosses the $AC$ side at points $ A$ and $P$, while $Γ_2$ crosses the AB side at points $B$ and $Q$. Let $X$ or line intersection point BP with $Γ_2$ so that $P$ is between $B$ and $X$. Likewise, be $Y$ the intersection point of the line $QC$ with $Γ_1$ so that $ Q $ is between $C$ and $Y$. Knowing that $A, X$ and $Y$ are collinear, calculate the smallest possible value to measure the $\angle{BAC}$ angle.

I think triangular inequality would help

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The triangle inscribed in $ABC$ with the least perimeter is the triangle, called the orthic triangle, whose vertices are the feet of the three altitudes of $ABC$.

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  • $\begingroup$ Sorry, but what does this imply? $\endgroup$
    – Lambert macuse
    Oct 19, 2019 at 17:25
  • $\begingroup$ I thought that this was what you wanted to know - "We want to find points M, N, and P on the sides BC, C A, and AB, respectively, such that the perimeter of MNP is minimal." $\endgroup$
    – user502266
    Oct 19, 2019 at 17:30
  • $\begingroup$ It was just an arbitrary idea to see if something came out of the problem. The real problem is calculating the possible Angle values $\endgroup$
    – Lambert macuse
    Oct 19, 2019 at 17:33
  • $\begingroup$ I think triangular inequality would help $\endgroup$
    – Lambert macuse
    Oct 20, 2019 at 10:31

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