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.

I'm not sure about this condition Problem wrote: ''Let ABC be a right triangle...'' If $A=90^{\circ}$ then $P=D$ or else $D=B$ or $C$.

  • $\begingroup$ could you add a figure. $\endgroup$
    – sirous
    Oct 21, 2019 at 15:48
  • $\begingroup$ Do not know :( .. $\endgroup$ Oct 21, 2019 at 18:20
  • $\begingroup$ Maybe my translation was not so good, I think you can understand. I put the test above $\endgroup$ Oct 22, 2019 at 0:34
  • $\begingroup$ May be you mean: Let $C_1$ and $C_2$ be the circles inscribed in triangles ABD and ADC. These circles are tangent to .... It is not clear, that is why I ask for a drawing. $\endgroup$
    – sirous
    Oct 22, 2019 at 17:17
  • $\begingroup$ I posted the original problem above, sorry if it was not so clear ... $\endgroup$ Oct 22, 2019 at 17:30

1 Answer 1


enter image description here

In this optimized drawing we have:



$\angle BPA=90^o$

$\angle BQY=90^o$

BC=a, AC=b, AB=c

In triangle ACX, AC is diameter of circle so $\angle CXA=90^o$ and we can write:


$AX \times YX=AX \times 2 AX=2AX^2$

$AX \times YX= b^2$

$QY \times QC = AX \times YX$

$QC=b Cos(\angle QCA)=b Sin(\angle CAQ)$


$CY \times QY =a(a- CQ)=a(a-b Sin (\angle CAQ))=b^2$

Let $\angle CAQ=\angle CAB=\alpha$

$Sin (\alpha)= \frac{a^2-b^2}{ab}=\frac{a}{b}-\frac{b}{a}$ . . . . . . . . (1)

In triangle ABC we can write:

$Sin (\angle CBA)=Sin 2(PBA)=2Sin(90-CAB)Cos(90-CAB))= 2 Sin(CAB)Cos(CAB)=Sin 2(CAB)$

$\frac{b}{Sin (\angle CBA)}=\frac{a}{Sin (\alpha)}$

$\frac{b}{Sin 2(CAB)}=\frac{a}{Sin (CAB)}$

$\frac{b}{2 Cos (\alpha)}=a$

$\frac{b}{a}=2 Cos (\alpha)$

Putting this in relation (1) we get:

$Sin (\alpha)=\frac{1}{2 Cos(\alpha)}-2 Cos (\alpha)= \frac{1- 4 Cos^2(\alpha)}{2 Cos(\alpha)}$

$2Cos^2 (\alpha)=1-4 Cos^2(\alpha)$


$Cos (\alpha)≈ 0.4$$\alpha ≈ 66^o $

  • $\begingroup$ Thank you so much for noticing me! Great resolution! $\endgroup$ Oct 28, 2019 at 21:19
  • $\begingroup$ you are welcome. $\endgroup$
    – sirous
    Oct 29, 2019 at 6:39

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