Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 793.
Monday, July 30, 2012
Problem 793: Right Triangle, Altitude, Incircles, Incenters, Circumcircle, Circumradius, Perpendicular, Distance, Metric Relations
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1)angle DBE=90/2=45 degrees
ReplyDelete2)points A,D,I,N are in the same line and this line is perpendicular to BE. This way the angle BDE is 45 degrees.
Similarly, points C,E,I,M are in the same line and this liner is perpendicular to BD. This way the angle BEI=45 degrees
3)bxc=2RxHa(b,c are triangle sides and Ha is altitude)
IDxIE=2RxIF
IDxIE=2IMxIN
IMxIN=RxIF, => R=(IMxIN)/IF.
Erina
Note that angle IDM = angle IEN = 45 degrees
ReplyDeleteR
= [ID*IE*DE] / [4*S(IDE)]
= [sqrt(2)*IM*sqrt(2)*IN*DE] / [4*S(IDE)]
= [IM*IN*DE] / [2*S(IDE)]
= IM*IN/IF
Let IK is perpendicular to AC. From similarity of IKM and INF, we have that IK/IM=IN/IF or R/IM=IN/IF.
ReplyDeleteTrue or false?
ReplyDelete(i)The incircle of ∆ABC, the circmcircle of ∆DEI and the circumcircle of ∆BDE are of equal radii
To Pravin: It is true. See problem 794
DeleteLet P,Q be points where BM and BN extended intersect with AC
ReplyDeleteAs derived earlier in Pr. 792
P,D,I,E,Q are concyclic wit PQ being the diameter. Also PQ=2R (R being the inradius of ABC and circum radius of DIE). It is also derived that I is the circumcenter of PBQ.
=>m(IED)=m(IPD)=m(IBP)=A/2 and m(IDE)=C/2
DIE and IPD are both C/2-135-A/2 triangles and from ASA both are congruent
=> DE=PI
As PIQ is an isosceles Right Tr. => PI=Sqrt(2). R=DE----------------(1)
Simiarly DME is similar to IFE
=>DM.IE=IF.DE
But DM=IM and IE=Sqrt(2).IN
=>IM.Sqrt(2).IN=IF.Sqrt(2)R
=>R=IM.IN/IF