Awesome Tangents!

Geometry Level 2

Circles of radii 3,4, and 5 units are externally tangent. The lines which form the 3 common external tangent intersect at $P,$ which is equidistant from the 3 points of tangency. Find this distance (from $P$ to any point of tangency)?

The answer is 2.236.

2 solutions

Sreekanth Narayan
Oct 28, 2014

The tangent is perpendicular to the radius at each points of contact and hence we will get a triangle of sides 7,8,9 on joining the centers. Since the point P is equidistant from the points of contact, the problem reduces to finding the inradius of the aforementioned triangle.

The area is $12\sqrt 5$ and since area equals semiperimeter times inradius, the inradius is $\sqrt{5} \approx 2.236.$

Exactly how I did it! Well done for the nice solution.

Michael Ng - 6 years, 7 months ago

Same wayyy!!

math man - 6 years, 7 months ago

Nice solution!!

Divyanshu Chandanan - 5 years ago

I solved the problem the exact same way, but can someone explain or prove why P is the in-center of triangle ABC in the first place? In-center is the intersection of angle bisectors but none of the tangent lines passes through the vertices of triangle ABC and so cannot be angle bisectors.

Sophie Ho - 4 months, 2 weeks ago

Wouter Dobbelaere
Nov 11, 2020

I used basic cos and tan rules to calculate the same easily (no need to know formula of inscribed circle...)

Use cos rule to find angle in C in triangle ABC: cos(C)=2/7 or C=48.2°

Then use tan rule in triangle PCF to calculate the requested distance:

5.tan(C/2) =2.236

Awesome Tangents!

The answer is 2.236.

2 solutions

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