Coxeter Exercise 4 of Section 1.3 in "Introduction To Geometry"

In this note we do the exercise listed in the title according to the hint given by the author.

   In the last blog entry it was shown that right triangles BOR and RPP1 in the diagram above were similar by constructing a circle whose center was the midpoint of segment OB, which included points B, P, O, R, and thus had inscribed angles BPR and BOR both subtended by chord BR and therefore equal to each other. We will call this circle [BPOR]. This result is the key to the exercise which is to prove the Erdos-Mordell Theorem: If O is any point inside a triangle ABC and P, Q, R are the feet of the perpendiculars from O upon the respective sides BC, CA, AB, then OA + OB + OC ≥ 2(OP + OQ + OR).
   The hint supplied by Coxeter is this: Let P1 and P2 be the feet of perpendiculars from R and Q upon BC. Define analogous points Q1 and Q2, R1 and R2 on the other sides. Using the similarity of the triangles PRP1 and OBR, express P1P in terms of RP, OR, and OB. After substituting such expressions into OA + OB + OC ≥ OA(P1P + PP2)/RQ + OB(Q1Q + QQ2)/PR + OC(R1R + RR2)/QP, collect the terms involving OP, OQ, OR, respectively.
   In the diagram above the points P1 and P2 are shown along with the feet of the perpendiculars from the arbitrary point O in ABC. Including the perpendiculars from P to Q1, R to Q2, P to R1, and Q to R2 makes for a pretty complicated drawing. It is suggested the reader draw the other two cases separately in order to better picture the argument below. Note: To play around with the parameters of the diagram, try different (sliding) values of the variables defined at the left in this diagram.
   Just as we constructed circle [BPOR] and established the similarity of triangles BOR and RPP1, we can construct five other such circles and pairs of similar triangles. Thus, circle [CPOQ] yields the similarity of triangles COQ and QPP2, circle [CQOP] yields COP similar to PQQ1, circle [AQOR] yields AOR similar to RQQ2, circle [BROP] yields BOP similar to PRR1, and circle [AROQ] yields AOQ similar to QRR2.
   From these similarities we can write:
PP1/OR = RP1/BR = RP/OB, so PP1 = (OR x RP)/OB
QP2/CG = PP2/OQ=PQ/OC, so PP2 = (OQ x PQ)/OC
PQ1/PC = QQ1/OP = PQ/OC, so QQ1 = (OP x PQ)/OC
RQ2/RA = QQ2/OR = RQ/OA, so QQ2 = (OR x RQ)/OA
PR1/BP = RR1/OP = RP/OB, so RR1 = (OP x RP)/OB
QR2/AQ = RR2/OQ = RQ/OA, so RR2 = (OQ x RQ)/OA.
   Also notice that in the quadrilateral RP1P2Q, the distance between parallel segments RP1 and QP2 is P1P2 and therefore RQ ≥ P1P2 = P1P + PP2, or (P1P + PP2)/RQ ≤ 1. Analogously, we get
(Q1Q + QQ2)/PR ≤ 1 and (R1R + RR2)/QP ≤ 1. Thus,
OA + OB + OC ≥ OA (P1P + PP2)/RQ + OB (Q1Q + QQ2)/PR + OC (R1R + RR2)/QP.
   Now substitute the expressions obtained by the similarities into the right hand side of this last expression and collect terms to get
OA + OB + OC ≥ OR[OA.RP/RQ.OB + OB.RQ/RP.OA] + OQ[OA.PQ/RQ.OC + OC.RQ/PQ.OA] + OP[OB.PQ/RP.OC + OC.RP/PQ.OB] which is of the form
OA + OB + OC ≥ OR[z + 1/z] + OQ[y + 1/y] + OP[x + 1/x], where x, y, z > 0.
   If f(u) = u + 1/u, u > 0, then f'(u) = 1 - 1/u^2 = 0 when u=1, f''(1) > 0, so the minimum value of f is f(1) = 2. Thus OA + OB + OC ≥ 2(OR + OQ + OP).
 

Similar Triangles

     Two triangles are called similar when they have the same three angles. When triangles are similar they have the same shape. That is, corresponding sides (the sides opposite equal angles) are in proportion. So if triangle ABC is similar to triangle DEF where angle A = angle D, angle B = angle E, and angle C = angle F, then BC/EF = AC/DF = AB/DE, where, for example, AB is the length of the side from point A to point B in triangle ABC, etc.
   In general, since the angles of a triangle always add up to 180 degrees, we only have to show two of the angles are equal because that implies the remaining ones are equal. In the case of two right triangles, since one angle is known to be 90 degrees, we only have to show one of the other two angles are equal to establish similarity.
   In high school Geometry it is usually pretty straightforward to establish similarity in the problems one is asked to solve. However, there are some cases where things get a little complicated. This note is about one such case.
   In the following diagram we have OR perpendicular to BR, OP perpendicular to BP and RP1 perpendicular to BP at point P1. We want to prove that triangle OBR is similar to triangle PRP1. Since they are both right triangles it is sufficient to show that angle ROB is equal to either angle PRP1 or angle RPP1. From the diagram it "looks like" it should be the latter. Before proving this result I would encourage the reader to play around with this diagram to try to establish the equality by straightforward methods. After all, with all these right triangles around it seems one could deduce this result without too much trouble. When I tried this exercise I was frustrated after many such straightforward attempts and a restless night of sleep. Eventually, an idea popped into my head that yielded the desired result. The fact that it wasn't so straightforward raises an interesting question in itself. That is, what makes a relatively simple problem harder than it first appears? Or, how "obvious" are geometric relationships? Or, why are some things harder to prove than others? Thoughts for another day.

NOTE:    This problem arose as part of Exercise 4 in Section 1.3 of H.S.M. Coxeter's "Introduction to Geometry". The problem in Coxeter is to prove the Erdos-Mordell theorem: If O is any point inside a triangle ABC and P, Q, R are the feet of the perpendiculars from O upon the respective sides BC, CA, AB, then OA + OB + OC ≥ 2(OP + OQ + OR). [The hint he gives is to let P1 and P2 be the feet of the perpendiculars from R and Q upon BC and to define analogous points Q1 and Q2, R1 and R2 on the other sides. Then, using the similarity of the triangles PRP1 and OBR, P1P can be expressed in terms of RP, OR, and OB. This can be done for the other analogous similar triangle pairs. Finally, noting that RQ ≤ P1P + PP2, etc., we can write OA+OB+OC ≤ OA(PP+PP2)/RQ + OB(Q1Q +Q2)/PR + OC(R1R+RR2)/QP. Finally, substitute expressions and collect terms involving OP, OQ, OR to get desired result.] So everything depends on establishing the similarity of the triangles PRP1 and OBR in the diagram above.

   The idea that opened the way for me was to notice that the angles RPP1 and ROB are both subtended by the segment BR. Thus, if a point could be found that was equidistant from B, R, O and P then we could draw a circle containing these four points which contained the chord BR subtending  inscribed angles RPP1 and ROB thus making them equal. Of course, it is not always true that given any four points such a circle can be found, but in this case it did work out. Let's see how to find the center of such a circle.
   Since the center must be equidistant from points R and O it lies on the perpendicular bisector of segment RO. Let D and E be the points where this perpendicular bisector of RO meet segments BO and RO respectively. Now look at triangle RDO with perpendicular bisector DE. Since OE=ER, ED is common, and the bisector is perpendicular to RO, the two triangles formed by the bisector are congruent (SAS). Thus RD = OD. Now we look at the angles. Angle ROB is the complement of angle RBO and it is also equal to angle ORD which is the complement of angle ORD. Thus angle RBO = angle DRB so triangle DRB is isosceles and DR = DB. Thus, B, R, and O are equidistant from D.
   Now we can make the same argument with respect to O and P with perpendicular bisector D'F of OP meeting BO at D' and OP at F. From the analogous argument we get D' equidistant from B, P and O.
   Now we need to show that D=D'. But we know that D and D' are both on segment BO and that both are the midpoint of BO (DO=DR=DB from the first argument and D'O=D'P=D'B from the second) so D=D' and we will call this common point D the center of the circle of radius DB that includes points B, P, O, and R.
   Finally, since chord BR subtends both angles RPB and ROB in this circle, we know these angles are equal and so the triangles BOR and RPP1 are similar. QED

   The amazing thing is that this result holds for any point O inside any triangle (as described in the NOTE above). The complete exercise will be done in the next blog entry coming soon.