triangles for Olympiad

                     circles

       geometry miscellaneous 

properties of triangle 

Let ABC be a triangle with sides 51, 52, 53. Let D denote the incircle of triangle ABC. Draw tangents to D which are parallel to the sides ABC. let r1. r2, r3 be the inradii of the three corner triangles so formed, find the largest integer that does not exceed r1+r2+r3. PRMO 2019

ANSWER=15

Let PQ be one of tangents parallel to BC and meet sides AB and AC at P and Q let PQ=x and BC=51

triangle ABC similar with triangle APQ thenxa=r1r=s−as which is in same way for yb and zc then r1+r2+r3r=xa+yb+zc=3-2=1

then r1+r2+r3=r and r by given condition of question =(s(s−a)(s−b)(s−c)s)12=(78(78−51)(78−52)(78−53)78)12=15.

2. In ∆ABC If , the straight lines AD, BE, CF are drawn through a point P to meet BC, CA, AB at D, E, F”  PROVE THAT   PD/AD+PE/BE+PF/C=1 and AP/AD+BP/BE+CP/CF=2     respectively?

3)Six congruent isosceles triangles have been put together as described in the picture below. Prove that points M, F, C lie on one line.( Israel MO 2019 )

Solution 

You need to show that M, F, C lie on a straight line. Observe that it can be shown that they are collinear if we can show that ∠EFM=∠CFD

Let’s investigate the triangle FDC.

 Observe that EF = AD and AD = AC.

 This results in the fact that FDC is

 isosceles

 and ∠FDC=π–∠EDA.

Now, we will try to infer 

something about triangle MEF. 

Observe that KM || HI as 

∠MKJ=∠KJH.

 Hence KHIM must be a parallelogram.

 Hence, KH || MI. Also, ∠KHI=∠HIG. 

 Hence, KH || EG. 

 Hence, it implies from  KH || MI 

and KH || EG, that M,E,I,G are collinear.

Now, this diagram ends it all. Observe that MI = KH. Also, EI = LJ. Hence, ME = KL = EP. Hence, MEF is isosceles.  Also, 

∠MEF=π–∠GEF=π–∠EDA=∠FDC. Hence, triangle MEF is similar to triangle FDC. This implies that ∠EFM=∠DFC.

4)Find alpha 

Solution 

5)In the diagram below, △ABC and △CDE are two right-angled triangles with AC = 24, CE =7 and ∠ ACB = ∠ CED. Find the length of the line segment AE 

Solution 

Let ∠ACB=∠CED=θ$\mathrm{\angle }ACB=\mathrm{\angle }CED=\theta$. That means that ∠ECD=90∘−θ$\mathrm{\angle }ECD={90}^{\circ }-\theta$ by the angle sum of △CDE$\mathrm{△}CDE$.