ABCD is a cyclic quadrilateral such that the diagonals AC and BD are perpendicular and their intersection is P. Point Q on the segment CP is such that CQ=AP. Prove that the perimeter of triangle BDQ is at least 2AC. Tuymaada 2017 Q2 Juniors
Problem
Source: Tuymaada 2017 Junior Level
Tags: geometry, inequalities
17.07.2017 11:56
Set coordinates of points as A,B,C,D=(−a,0),(b,0),(0,c),(−d,0) where a,b,c,d>0 and P=(0,0). Then Q=(c−a,0) Perimeter of BDQ =√b2+(c−a)2+√(c−a)2+d2+(b+d)≥√(b+d)2+4(c−a)2+(b+d) We will prove, that √(b+d)2+2(c−a)2+(b+d)≥2AC=2(a+c) or √(b+d)2+2(c−a)2≥2(a+c)−(b+d) 4(c−a)2≥4(a+c)2−4(a+c)(b+d) (a+c)(b+d)≥4ac As ABCD is cyclic, then ba=cd=tanBAC=x So (a+c)(b+d)=(a+c)(ax+cx)≥4ac
17.07.2017 15:17
can anybody post other questions?
17.07.2017 21:25
lemma Let B,C,U,V 4 points s.t. U is on the BC bisector and UV∥BC then BV+VC≥BU+UC proof : let C′ the symmetric of C in UV then B,U,C′ are collinear and BV+VC=BV+VC′≥BC′=BU+UC′=BU+UC Back to the problem Let O,I the circumcenter of ABCD, the midpoint of AC ;B′,D′the symmetrics of B,D in I BQ+BP≥BB′,DP+DQ≥DD′ so the perimeter ≥2(BI+ID)by Lemma≥2(BO+OD) since IO∥BD therefore we get the desired result
30.04.2018 17:36
Let's note PQ=x , AP=QC=a , BP=b and DP=c. Because ABCD is cyclic we have BP∗PD=AP∗PC , so bc=a(x+a).The perimeter of triangle BDQ is √b2+x2+√c2+x2+b+c.But we have a2+ax−bc=0 , so a=−x±√x2+4bc2 . Because a is the length of AP , a is a positive number , so a=−x+√x2+4bc2. We must prove this inequality√b2+x2+√c2+x2+b+c>=2(2a+x) which is equivalent to √b2+x2+√c2+x2+b+c>=√x2+4bc . By Minkowski inequality we have √(b+c)2+4x2+b+c>=2√x2+4bc Square it and after divide by 2 we obtain (b+c)2+(b+c)√(b+c)2+4x2>=8bc which is true because (b+c)2>=4bc and (b+c)√(b+c)2+4x2>=(b+c)(b+c)>=4bc. Equality when x=0 , P=Q so when ABCD is a square.
28.12.2018 13:01
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