# Calculate the smallest possible value to measure the ∠BAC angle

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 $$C$$.

• could you add a figure. Oct 21, 2019 at 15:48
• Do not know :( .. Oct 21, 2019 at 18:20
• Maybe my translation was not so good, I think you can understand. I put the test above Oct 22, 2019 at 0:34
• 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. Oct 22, 2019 at 17:17
• I posted the original problem above, sorry if it was not so clear ... Oct 22, 2019 at 17:30 In this optimized drawing we have:

BA=BC=CY

AX=AY

$$\angle BPA=90^o$$

$$\angle BQY=90^o$$

BC=a, AC=b, AB=c

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

$$2AX^2=AC^2=b^2$$

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

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

$$QY \times QC = AX \times YX$$

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

$$CY=AB=BC=a$$

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

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:

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

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

$$Cos^2(\alpha)=\frac{1}{6}$$

$$Cos (\alpha)≈ 0.4$$$$\alpha ≈ 66^o$$

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