Minnie 3rd Birthday Wording

Positions

Statement

< j ≤ 2i.

Crossover ^{Make funny drawing correctly is crucial to solving a problem of this type. You}

from a triangle ABC which has no specific regularity, because a isosceles or equilateral different centers can mix and confuse the solution. Nor should we use for Q the center of the side. As we need the incenter of the two triangles ACQ and BCQ, it is logical to draw two of the bisectors from Q, as it saves labor. Of course, as AQC and BQC add up to 180 degrees, Clearly, both bisectors form a 90 degree angle.

^{Thus, segments} QI 1 and QI ^{2 form an angle of 90 degrees, so that the triangle QI} I 1 2

is rectangle Q. Since the hypotenuse is the diameter of the circle we have built, this means that Q belongs to the circle (at the beginning of the arc enrolled).

incircle

To follow reasoning, they should work a bit with the inscribed circles. There are several properties that we know. If you look at the second picture, we call these points At tangency (segment BC), Bt (segment BC) and Ct (segment AB), that BC = BAt + CAt, AB = ACt + BCT and CBT + AC = Abt. We note that the distances to the vertices of these points of contact are equal by symmetry, so that ABT = ACt, BAT and BCT = CAt = CBT. In addition, the perimeter of the triangle is 2 * ACt + 2 * + 2 * CAt BCT, so that p is semiperimeter ACt + BCT + CAt. In this way, you have to ACt = p - BC (because BAt + CAt = BC), BCT = p - AB and CAt = p - AB. In our example, if we place the points of contact of the two circles in the drawing, it is easy to reason for similarities in the segment QP, the two points of tangency are equidistant from the ends, so QP is the sum of the distances from Q to the points of tangency of two circles. According

property we have seen before in the triangle ACQ distance Q to the tangent point is its semiperimeter less side BC and the triangle BCQ, the distance is the least semiperimeter BC. If we add the two distances for QP, we have the sum of the two semiperimeter less AC and BC. By adding the two semiperimeter get the original triangle semiperimeter more QC, ie QC QP = p + - AC - BC = QC + p - 2p + AB = QC - p + AB.

Therefore, the distance from P to C QC - QP = P - AB which is what I wanted to prove.