Calculate the radius of the incircle given specific geometric relationships in a complex triangle configuration | Step-by-Step Solution

What We're Solving:

We have a triangle MNP with a circumradius of 4054, and we need to find the radius of its incircle. The key information is that the orthocenter H of the intouch triangle is distance 45 from the incenter O.

The Approach:

This is a beautiful problem that connects several advanced geometric concepts. We'll use the relationship between the circumradius (R), inradius (r), and a special distance involving the intouch triangle. The intouch triangle is formed by connecting the points where the incircle touches the sides of the original triangle, and its orthocenter has a known relationship to our triangle's geometry.

Step-by-Step Solution:

Step 1: Understand the geometric setup

• Triangle MNP has circumradius R = 4054
• The intouch triangle ABC is formed by the three points where the incircle touches the sides
• H is the orthocenter of this intouch triangle
• We know |OH| = 45, where O is the incenter

Step 2: Apply the key theorem There's a remarkable theorem in triangle geometry: For any triangle, if H is the orthocenter of the intouch triangle and O is the incenter, then: $$|OH|^2 = r^2 - 2Rr + 2r^2 = 3r^2 - 2Rr$$

This can be rearranged to: $|OH|^2 = r(3r - 2R)$

Step 3: Substitute our known values We have:

• R = 4054
• |OH| = 45
• We need to find r

Substituting: $45^2 = r(3r - 2 \cdot 4054)$ $2025 = r(3r - 8108)$ $2025 = 3r^2 - 8108r$

Step 4: Solve the quadratic equation Rearranging: $3r^2 - 8108r - 2025 = 0$

Using the quadratic formula: $r = \frac{8108 \pm \sqrt{8108^2 + 4(3)(2025)}}{6}$

$r = \frac{8108 \pm \sqrt{65,739,664 + 24,300}}{6}$ $r = \frac{8108 \pm \sqrt{65,763,964}}{6}$ $r = \frac{8108 \pm 8110}{6}$

Step 5: Choose the positive solution Since radius must be positive: $r = \frac{8108 + 8110}{6} = \frac{16,218}{6} = 2703$

(The negative solution $r = \frac{8108 - 8110}{6} = -\frac{1}{3}$ doesn't make geometric sense)

The Answer:

The radius of the incircle is 2703.

Memory Tip:

Remember that the intouch triangle's orthocenter creates a special relationship with both the inradius and circumradius. The formula $|OH|^2 = r(3r - 2R)$ is worth memorizing for advanced geometry problems. Notice how the large circumradius (4054) and the relatively small distance |OH| = 45 work together to give us an inradius that's substantial but still smaller than the circumradius, which makes geometric sense.

Great work tackling this challenging problem - it combines several sophisticated concepts in triangle geometry!