Question

Let ∆ ABC be equilateral. On side AB produced, we choose a point P such that A lies between P and B. We now denote α as the lengths of sides of ∆ ABC; r₁ as the radius of incircle of ∆ PAC and r₂ as the ex radius of ∆ PBC with respect to side BC. Then prove that r₁ + r₂ equals \(\frac{α\sqrt{3}}{2}\) .

Answer 1

Let O₁ be the centre of the in-circle of ∆ PAC and O₂ the centre of the circle which touches the triangle PBC on side BC. Let the tangents from P on these two circles touch them at points T₁, T₁′ and T₂, T₂′ respectively. 

Looking at the figure, we see that ∠T₁O₁R = 60° since each of ∠s O₁T₁A and O₁RA being = 90°, it is the supplement of ∠T₁AR = 120° (as an exterior angle for ∆ ABC). Hence, ∠AO₁R = 30°. Similarly, we obtain ∠BO₂S = 30°.

Since tangents drawn to a circle from an external point are equal, we have

T₁ T₂ = T₁ A + AB + BT₂ = RA + AB + SB 

= r₁ tan 30° + α + r₂ tan 30° = \(\frac{r_1 +r_2}{\sqrt{3}}\) + α

and 

T'₁ T'₂ = T'₁ C + CT'₂ = CR + CS = (α – RA) + (α – SB)

= 2α - \(\frac{r_1 +r_2}{\sqrt{3}}\) 

Since common external tangents to two circles are equal, T₁ T₂ = T'₁T'₂. Hence,

 \(\frac{r_1 +r_2}{\sqrt{3}}\) + α =  2α - \(\frac{r_1 +r_2}{\sqrt{3}}\)  

Hence we find that, r₁ + r₂ = \(\frac{α\sqrt{3}}{2}\)