Geometry (circle inscribed into triangle)

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Discussion Overview

The discussion revolves around finding the radius of a circle inscribed in a triangle, with participants exploring various methods and reasoning related to geometric properties and relationships within the triangle.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a proposed answer for the radius as \(\frac{2\sqrt{5}}{3}\) but seeks an explanation for this result.
  • Another participant suggests a method involving drawing perpendicular bisectors from the triangle's edges to the center of the inscribed circle, indicating that the radius can be derived from the areas of the smaller triangles formed.
  • A different approach is introduced, where a participant claims that the radius perpendicular to the base bisects the base, leading to a relationship involving the lengths of the sides and the Pythagorean theorem.
  • Another solution is proposed using two congruent triangles, establishing a ratio involving the segments created by the radius and the triangle's dimensions.
  • One participant questions how certain values for the segments created by the radius tangent to the longer side are determined, specifically referencing segments of lengths 5 and 2.
  • A repeated question seeks clarification on the reasoning behind the segment lengths, suggesting that the smallest segment is half of the triangle's base.

Areas of Agreement / Disagreement

Participants present multiple competing methods and reasoning for finding the radius, indicating that there is no consensus on a single approach or solution. The discussion remains unresolved with various interpretations and calculations being explored.

Contextual Notes

Some methods rely on specific geometric properties and relationships that may not be universally accepted or understood, leading to potential gaps in assumptions or definitions. The discussion includes various mathematical steps that are not fully resolved.

Inspector Gadget
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I'm working on this problem.

The only given information is the circle is inscribed into the triangle. And you have to find the radius of the circle.

The answer is...

\frac{2\sqrt{5}}{3}

Can someone come up with an explanation as to why? It's been a few years since I had geometry, and have tried everything I remembered with bisectors and whatnot.
 
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Here's one way, there are probably several: draw in the perpendicular bisectors from the edge of the triangle to the centre of the circle (ie starting from the points of contact with the circle). The radius is then the altitude of the three new triangles so defined. Ues their areas to work out the area of the large triangle, which you can in turn find using only the lengths of the sides. Benefit of this method: no sin or cosines to work out.
 
My solution: We can prove that the radius perpendicular to the base also bisects the base (it is on the same line as the perpendicular bisector). So, the base equals 2. The entire perpendicular bisector equals (45)^(1/2) (7^2-2^2=perp^2). Now, we know that the longer sides will also have a tangent radius. This radius will split it into two pieces, one of length 2 and the other of length 5. The piece of length 5 has relation to the perp bisector - the radius. Using the equation of pythagoras, 5^2+r^2=((45)^(1/2)-r)^2
Simplifying will give you your answer.
 
Another solution: using two congruent triangles:
T1. Top vertex (A), base midpoint (B), right vertex (C),
T2. Top vertex (A), circle's center (D), intersection between the circle and AC (call this intersection E).

Then, DE/EA = CB/BA.

And you know that: DE is the radius, EA=7-2=5, CB=2, BA=sqrt(45). Solve for r (or DE), substitute, done.
 
Quick question on an old thread, how do Wooh and ahrkron know that the radius tangent to the longer side will split the side into segments of 5 and 2?
 
Genza said:
Quick question on an old thread, how do Wooh and ahrkron know that the radius tangent to the longer side will split the side into segments of 5 and 2?

The smallest segment equals the half of the triangle's base.
 

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