E-320: Teaching Math with a Historical Perspective Oliver Knill, 2010Lecture 3: special points for triangles1. ObjectiveSome coincidences which happen in triangles are usually taken for granted because they are ratherelementary. Still, it is important to appreciate that they are not so obvious and also to see whythey are true. Also terrific textbo oks in geometry do not bother with such things. Either theyassume them to be known or they assume them to be too trivial to bother with. But these factsare not trivialities, they are 4 b eautiful miracles. You can appreciate it only to a full extentwhen running the Mathematica code provided here and see how complicated the formulas for theintersection points can become, if the points are given in terms of general variables.1. The CentroidThe centroid of a triangle is the intersection of the lines which connect the vertices of a trianglewith the midpoints of the opposite side. Note that it is not at all trivial t hat these three linesintersect in one point. It is a stability property of the t riangle. D eforming a triang le does notchange this property. If A, B, C are the coordinates of the vertices, then ( A + B)/2, (A + C)/2and (B + C)/2 are the midpoints of the sides. To verify the property just check that withP = (A + B + C)/3, the points A, P, (B + C)/2 are on a line, the points (B, P, (A + C)/2 are ona line and t he points (C, P, (A + B)/2 are on a line. There is an easier but more advanced wayto see this: check it first for the equilateral triangle. Now, any triangle can be mapped into anyother by a linear transformation. Because linear transformations preserve lines and ratios, theintersection property will stay true for all triangles.Figure 1: To the left, we see the situation as we would expect it without ”knowing” that the threeintersection points agree. To the right, we wee the actual situation.Here is a proof using Mathematica. If you run the two lines, it should give the result ”True” andverify that the centroid is on the line segments between vertices and opposite midpoints and alsoshow that the lengths are divided in a ratio 2/1:a={aa , a2 , a2 } ; b={b1 , b2 , b3 } ; c={c1 , c2 , c3 } ; u=(b+c ) / 2 ; v=(a+c ) / 2 ;w=(a+b ) / 2 ;p=(a+b+c ) / 3 ; Simplify [ p == a/3+2u/3== b/3+ 2v/3 == c /3+2w/ 3 ] 2. The OrthocenterThe orthocenter is the intersection of the three altitudes of a triangle. Also here - a priory -we have three different points the intersection, for each pair of altitudes. Why do they meet inone point? It is not obvious and was not proven by the Greeks for example. One can take theintersection of two altitudes, get a point P and form the line from P to the third point in thetriangle. The fact that this line is perp endicular to the third line can be seen by looking at theangles. The angles between to heights is the same than the a ngle between the two correspondingsides.Figure 2: To the left, we see the situation as we would expect it without ”knowing” that the threeintersection points agree. To the right, we wee the actual situation.Here is a brute force Mathematica verification: it computes the pedal points and verifies that theintersection between the lines coincide:a={a1 , a2 } ; b={b1 , b2 } ; c={c1 , c2 } ;f [ { a , b , c }]: = a+ ( b−a ) ∗ ( ( c−a ) . ( b−a ) ) / ( ( b−a ) . ( b−a ) ) ;m=f [ { b , c , a } ] ; n=f [{ c , a , b } ] ; p=f [ { a , b , c } ] ;L [ x , y , z , w ]: = x+( t / . Solve [ x+t ( y−x )==z+s (w−z ) , { t , s } ] [ [ 1 ] ] ) ( y−x ) ;Simplify [ L [ a ,m, b , n]==L [ a ,m, c , p]==L [ b , n , c , p ] ] 3. The Center of the Circumscribed CircleAny circle which passes through two points A,B of a tria ngle lies o n the perpendicular bisectorof A and B. When moving a point M on t hat line and always drawing the circle centered at Mthrough A,B, then there will be a moment, where the distance to the third point C is equal to thedistance to A. We have found the circumscribed circle of the triangle. The point can be obt ainedby taking the intersection of the three perpendicular bisectors.Figure 3: To the left, we see the situation as we would expect it without ”knowing” that the threeintersection points agree. To the right, we wee the actual situation.Here is a Mathematica verification which provides general formulas for the perpendicular bisectorsand their intersection.a={a1 , a2 } ; b={b1 , b2 } ; c={c1 , c2 } ; u1=(b+c ) / 2 ; v1=(a+c ) / 2 ; w1=(a+b ) / 2 ;p [ x ]:={−x [ [ 2 ] ] , x [ [ 1 ] ] } ; u2=u1+p [ c−b ] ; v2=v1+p [ c−a ] ; w2=w1+p [ a−b ] ;L [ x , y , z , w ]: = x+( t / . Solve [ x+t ( y−x )==z+s (w−z ) , { t , s } ] [ [ 1 ] ] ) ( y−x ) ;Simplify [ L [ u1 , u2 , v1 , v2]==L [ u1 , u2 , w1 , w2]==L[ v1 , v2 , w1 , w2 ] ] 4. The Center of the IncircleAny circle which is tangent to two sides of a triangle lies on the angle bisector at the intersectionpoint o f the sides. Take a circle on that line which is tangent to the two sides. If the center isclose to the point then the circle is small and inside the triangle. Move the point along the line.There will be a moment, when the circle will touch the third side. This point is the intersectionpoint of all angular bisectors. It is the center of the inscribed circle. The inscribed circle is thecircumscribed circle of the pedal points.Figure 4: To the left, we see the situation as we would expect it without ”knowing” that the threeintersection points agree. To the right, we wee the actual situation.Here is a proof using Mathematica. Running it gives the result ”True”. It constructs the anglebisectors by constructing the line through the triangle vertex point A and the midpo int and thepoints on the lines in distance 1 from the point.a={a1 , a2 } ; b={b1 , b2 } ; c={c1 , c2 } ;f [ a , b , c ] := a+ ( ( b−a )/ Sqrt [ ( b−a ) . ( b−a ) ] +(c−a )/ Sqrt [ ( c−a ) . ( c−a ) ] ) / 2u=f [ …