Fagnano's Problem: What is it?
A Mathematical Droodle

Izvolsky's solution


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Explanation

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The applet attempts to illustrate a relatively novel solution to Fagnano's problem. The solution by N. A. Izvolsky has been published in Russian in 1937 in the first series of collections Mathematics Education (n 10) and later included into S. I. Zeitel's New Triangle Geometry (UchPedGiz, 1962), also in Russian.


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Let O be the circumcenter of ΔABC and points P, Q, R located on sides BC, AC, and AB respectively. Joining O to P, Q, R yields three quadrilaterals: OQAR, ORBP, OPCQ. Let's start with the latter. The quadrilateral OPCQ consists of two triangles OCP and OCQ with common base OC. If h and g denote altitudes to the base in the two triangles, we can write

 
Area(OPCQ)= Area(OCP) + Area(OCQ)
 = h·OC/2 + g·OC/2
 = (h + g)·OC/2
 ≤ PQ·OC/2.
 ≤ PQ·r/2,

where r is the circumradius of ΔABC. Similarly we obtain

 
Area(OQAR)≤ QR·r/2 and
Area(ORBP)≤ PR·r/2.

Adding the three we see that

 
Area(ABC)= Area(OQAR) + Area(ORBP) + Area(OPCQ)
 ≤ QR·r/2 + PQ·r/2 +PR ·r/2
 ≤ (QR + PQ + PR)·r/2.

Hence

(1) Perimeter(PQR) ≥ 2·Area(ABC)/r.

Now, by Nagel's theorem, the sides of the orthic triangle are perpendicular to the radius-vectors from O to vertices of ΔABC. Which means that, for the orthic triangle, (1) turns in an identity. Thus the orthic triangle solves Fagnano's problem. That the solution is unique follows from the fact that for any inscribed triangle PQR (1) becomes an identity if and only if its sides are perpendicular to the said radius-vectors. The sides of such a triangles are then parallel to the sides of the orthic triangle. But this is impossible for an inscribed triangle different from orthic.

Fagnano's problem

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