Math IOQM

The statement Arc-midpoint (perpendicular-bisector ∧ angle-bisector) lemma – often nick-named “Fact 5”. In any triangle A B C ABC , the intersection E E of the internal (or external) bisector of ∠ C \angle C , and the perpendicular bisector of A B AB lies on the circumcircle ( A B C ) (ABC) . In fact, E E is the midpoint of the arc A B AB : with the internal bisector it is the midpoint of the arc A B AB not containing C C ; with the external bisector it is the midpoint of the arc A B AB containing C C . Proof: 1. Mid-arc argument (the slick one) Draw the circumcircle ω = ( A B C ) \omega=(ABC) and let M M be the midpoint of the minor arc A B AB (the arc that does not contain C C ). Why M M is on the perpendicular bisector of A B AB . Equal arcs subtend equal chords, so M A = M B MA=MB . Hence M M lies on the perpendicular bisector of A B AB . Because perpendicular bisector of AB contains all the points equidistant from A,B. Why M M is on...

As long as the three points are not all on one straight line, there is exactly one circle that passes through them . Why it’s unique Perpendicular-bisector construction Draw the perpendicular bisector of the segment joining points A and B. Draw the perpendicular bisector of the segment joining points B and C. Because A, B, C are not collinear, those two bisectors intersect at a single point O. O is equidistant from A, B, and C (that’s how perpendicular bisectors work), so OA = OB = OC. Taking O as the center and OA (or OB or OC) as the radius gives a circle that goes through all three points. Uniqueness follows from basic geometry If some other circle also passed through A, B, and C, its center would have to be equidistant from those three points too, so it would lie at the same intersection of the two bisectors – exactly at O. Only one intersection means only one possible center, hence only one circle. Edge cases Collinear points : If A, B, C lie on the...

 

20. PS is a line segment of length 4 and O is the midpoint of PS. A semicircular arc is drawn with PS as diameter. Let X be the midpoint of this arc. Q and R are points on the arc PXS such that QR is parallel to PS and the semicircular arc drawn with QR as diameter is tangent to PS. How can I get the area of the region QXROQ bounded by the two semicircular arcs? Answer: 2*PI - 2 Solution: Image taken from here . Now, the big circle has area 4*Pi. Since RQ is the diameter of the smaller circle, it subtends 90 degree on its circumference. So Angle QOR is right angle. So the line segments QO and OR will capture 1/4th the area of the completed blue circle which is Pi.--------[1] Now we just need to find the area between green  lines and the red circular arcs below them. If you see the area captured by line segments YQ and YO in the red circle, it will be 1/4th of the completed red circle. Again the same reason that angle QYQ is right angle. Radius of the red circle is sqrt(2). So the c...

14. \( O \) and \( I \) are the circumcentre and incentre of \( \triangle ABC \) respectively. Suppose \( O \) lies in the interior of \( \triangle ABC \) and \( I \) lies on the circle passing through \( B, O, \) and \( C \). What is the magnitude of \( \angle BAC \) in degrees? You can apply a trick here. If you can construct a triangle which satisfies the conditions given in the question, you can answer using that. Here if you construct an equilateral triangle, O will lie in the interior of the triangle and I will lie on the circle passing through O,B,C since in the equilateral triangle O and I are same. So the answer will simply be 60 degrees. Detailed solution copied from here : We know, because \( I \) is the incenter, \[ \angle IBD = \frac{B}{2}, \quad \angle ICD = \frac{C}{2} \] \[ \Rightarrow \angle BIC = \pi - \left( \frac{B}{2} + \frac{C}{2} \right) = \pi - \left( \frac{\pi - A}{2} \right) = \frac{\pi}{2} + \frac{A}{2} \] But, \( \angle BIC = \angle BOC \), because c...

  Length of common tangent: Answer: 2.Sqrt(R.r) Answer: sqrt(r1.r3) Answer: 2 Answer: sqrt(c) = sqrt(ab)/sqrt(a) + sqrt(b) Answer: 12 Answer: 65 Answer: 70 Answer: 10 Answer: 7 Answer: 15,13 Answer: 15

  120, 30, 30 230 80,50 20.sqrt(3) Common tangents to 2 circles