## Saturday, March 2, 2019

### Geometry Problem 1419: Triangle, Altitude, Circumcircle, Incenter, Inradius, Perpendicular, Angle Bisector, Midpoint

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. 1. Fie P intersecția lui IF cu BH, T intersecția lui ED cu BH și V punctul de tangență al lui AC cu cercul I.
De aici rezultă că IV este perpendicular pe AC. Cum HF și HD sunt bisectoare rezultă triunghiurile HFP și HDE sunt dreptunghice isoscele, de unde HF = FP = FG și DH = DT = DE.
Din PF perpendicular pe LH rezultă LF = FH și de aici ușor LGHP pătrat, adică LG = GH.
La fel din TD perpendicular pe HN rezultă HD = DN și de aici NEHT pătrat, adică NE = HE.
Dar triunghiul GIE este dreptunghic isoscel și deci GV = VE = IV =r.
Acum LG + NE = GH + HE = GV + VE =2r.
Florin Popa , Comănești, România.

1. To Florin Popa

below is the translation of your solution to English by Google Translate . Hope that it translate correctly .
Let P be the intersection of IF with BH, T the intersection of ED with BH and V with the point of tangency of AC with circle I.
Hence, IV is perpendicular to AC. As HF and HD are bisecting, the HFP and HDE triangles are isosceles, where HF = FP = FG and DH = DT = DE.
From PF perpendicular to LH, LF = FH and LG LG square, LG = GH.
Likewise, TD perpendicular to HN results HD = DN and hence NEHT square, ie NE = HE.
But the GIE triangle is a right-angled isoscelet and therefore GV = VE = IV = r.
Now LG + NE = GH + HE = GV + VE = 2r.
Florin Popa, Comanesti, Romania.

1. per lines 4 and 5 of your solution "From PF perpendicular to LH, LF = FH and LG LG square, LG = GH.
Likewise, TD perpendicular to HN results HD = DN and hence NEHT square"
I am not sure if TD perpendicular to HN will results to HD = DN . Please explain.

2. Note that point N is the intersection of HD to the perpendicular bisector of EJ ( problem statement) was never been used in your solution. please explain

2. 1.Punctele T,I,D,E sunt coliniare (Un segment de linie prin I si perpendicular pe bisectoarea unghiului BHC la D intersecteaza HC la E).Diametru perpendicular pe coarda HN.
2.Nu a fost nevoie de perpendiculara pe Ej.( Florin Popa)

3. Lets assume in-circle touches AC at point Q, it is easy to see that GQ=QE=r,and GE=2r.
Let EI extended intersects BH at P and GI intersects BH at T.
We can use the results of Problem 1417, we have N as circumcenter of Tr.PJE and L as circumcentre of Tr.TKG.
HE and HP are tangent to Circle N, HG and HT are tangent to circle L. Quad. HENP and Quad. HGLT are squares, we have NE=HE and LG=HG, we get NE+LG=GE=2r.