ASSIGNMENT 8

An Investigation of Fagnano's Problem

by

Kwesi Yankey

Given a triangle, find the triangle of minimal perimeter that can be inscribed in it.

Consider the acute -angled triangle ABC in Figure 8.1

The problem is to find the triangle PQR such that it has minimal perimeter. The solution to this problem lies on the following premise:

Consider triangle LMN and an arbitrary point P as shown in Figure 8.2

Reflect P on LM to obtain P Thus LP= LP and <PLM=<MLP

Reflect P' on LN to obtain P" Thus LP'=LP"and <P' LN=<NLP"

Thus, we conlude that LP=LP =LP" and <PLP"=2<MLN.

We use the above result to find to the triangle of minimal perimeter.

In Figure 8.3,reflect PQ on side AB to obtain PQ. Similary, Similary, reflect PR on side AC to obtin PQ. Similary, reflect PR on side AC to obtain RP".The perimeter of triangle PQR is now P-P' Q+QR+RP"

For the perimeter to be minimized,P'QRP" should be collinear. Now, AP=AP=AP and <P' AP"=2<BAC. Triangle P' AP" is isosceles. Because <AP" is independent of the choice of P, the base P' AP is isosceles.

Because <P' AP" is independent of the choice of P the base P' P" will have minimum measure when the equal side AP' and AP" have minimal length.

However, these two sides are equal in length to AP and AP has minimum lenth when it is perpendicular to BC,i.e when AP is the altitude of triangle ABC at A. Similarly Q and R must be the feet of the other altitudes of triangle ABC.

Thus, the triangle with the minimum perimeter is the orthic triangle triangle (FIGURE 8.4).

The Orthic is the name given to the triangle joinning the feet of the altitudes at P, Q and some of its properties are

(1) Its respective angles can be computed by subtracting 2 times the opposite angles in the original triangle. For example in Figure 8.4,<QPR=180-2<PQR=180-2<ACB,and <QRP=180-<ABC.

(11) The altitudes of the original triangle ABC are the bisectors of the Orthic triangle PQR. Thus,the incenter of the Orthic triangle is the orthocenter of the original triangle.

(111) The perimeter of Orthic triangle is . Observe that this formula allows the computation of the perimeter of the Orthic triangle using the dimensions of the original triangle.

In Figure 8.4,AB=5.82cm, BC=7.83cm,CA=7.48cm PQ=3.20cm QR=2.60cm PR=4.13cm,<A=70.9,<B=64.5and <C44.6

Perimeter of Orthic triangle is L=BC cos A=ACcos B+AB cosC= 7.83cos 70.9+7.48cos 64.5+5.82cos 44.6=9.93cm

This is equal to PQ+QR+RP= 3.20+2.60+4.13=9.93cm.

(iv) Area of triangle ABC =,where L is the perimeter of the Orthic triangle and R is the radius of the Circumscribed circle of ABC.

In Figure 8.4 area of triangle ABC=Note that because the Orthic triangle depends on the altitudes of the original triangle ABC its vertices may be outside the triangle ABC.This occurs when ABC is an obtuse-triangle .For right-angled triangles ,the Orthic triangle degenerates to a line.