Four points A, B, C, and D lie on a straight line in the X-Y plane, such that AB = BC = CD, and the length of AB is 1 metre. An ant at A wants to reach a sugar particle at D. But there are insect repellents kept at points B and C. The ant would not go within one metre of any insect repellent. The minimum distance in metres the ant must traverse to reach the sugar particle is

Since ant cant go more near than 1 m.

So it'll have to travel in circular path from A till it is south of pt.

B and hence travels a quarter circle with radius = 1 and center as B, so travel $$\frac{\pi }{2}$$ m distance.

Then from B to C it travels 1m in straight line till C.

Then again from C to D in circular path way with distance $$\frac{\pi }{2}$$ m.

Hence total distance traveled = $$\frac{\pi }{2}$$ + $$\frac{\pi }{2}$$ +1 = $$\pi $$ +1.

Alternate solution : 

Let the given figure represent the situation. A,B,C,D are the points with distance between them 1 m. 

The repellents are at B and C respectively. The circles drawn with centre at B and C and radius equal to AB = 1 m. 

Let E and F be the points of intersection of the circles and G and H be the points on the circle perpendicular to B and C. 

Therefore, the ant can only travel at the circumference of the circles.

The shortest path for the to take will take is : A-G-H-D.

This is because taking the curve G-E and E-H will be a longer path than travelling straight from G to H. 

While travelling from G to H, the distance of the ant will be more than 1 m from the repellant. 

Thus, the distance travelled by the ant on the circumference of the circle ie 

arc AG = $$\dfrac{90}{360} \times 2 \times \pi 1 = \dfrac{\pi}{2} $$ 

Length of arc HD = length of arc AG = $$\dfrac{\pi}{2} $$ 

Total length = $$(2 \times \dfrac{\pi}{2})+ BC = \pi + 1$$