Proposed Problem
See complete Problem 25 at:
gogeometry.com/problem/p025_right_triangle_incircles_incenter.htm
Level: High School, SAT Prep, College geometry
Wednesday, April 8, 2009
Problem 25: Right triangle, Altitude, Incircles, Incenter, Angle Bisector
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Let incircles of Tr ABD & Tr.CBD meet AB & BC respectively in J & K. Now with the standard nomenclature we can show that: BD=ac/b,AD=c^2/b
ReplyDeleteHence, r1=Inradius of Tr.ABD=(ac/b*c^2/b)/(ca/b+c^2/b+c) = ac^2/(b(a+b+c))
Likewise, r2=Inradius of Tr.CBD=(ac/b*c^2/b)/(ca/b+c^2/b+c) = ca^2/(b(a+b+c))
Now take BA & Bc as the x-axis and y-axis. It can be shown by elementary geometry that F is
[ac^3/(b(b-a)(a+b+c)), ac^2/b(a+b+c)] since BJ is r1*cot(B/2). Likewise, G is [ca^2/b(a+b+c),ca^3/(b(b-c)(a+b+c))]. By the distance formula which is nothing but using Pythagoras: FG^2 = (ac)^2*(a^+c^2-ab)^2/(b(a-b)(a+b+c))^2 +ac)^2*(a^+c^2-bc)^2/(b(b-c)(a+b+c))^2
Now a^+c^2=b^2; thus, FG^2 = 2(ac/(a+b+c))^2
But r = Inradius of Tr. ABC = ac/(a+b+c)
Hence, FG^2 =2r^2 ---(1) while from Tr. BLE we've BE^2=BL^2+LE^2 where L is the point at which incircle of Tr. ABC touches AB. Also BL=LE=r. Thus, BE^2 =2r^2 ---(2)
By equations (1) & (2) we've, FG=BE=V2*r
Now slope FG = [ca^3/(b(b-c)(a+b+c)) - ca^2/b(a+b+c)]/[ca^2/b(a+b+c)-ac^3/(b(b-a)(a+b+c))]
= ac/(a+b+c)/ -[ac/(a+b+c)]= - 1 which means FG makes an angle of 135 deg with x-axis BA and since BE is the angle bisector of angle ABC (=90 deg) it makes an angle of 45 deg to the x-axis i.e. to BA which in turn means that BE is perpendicular to FG.
QED
Ajit: ajitathle@gmail.com
Let r, r1, and r2, are the inradii of the triangles ABC, ADB, and BDC respectively.
ReplyDeleteThen r=(AB*BC)/(AB+BC+CA),r1=(AD*BD)/(AB+BD+DA) and r2=(CD*BD)/(BC+CD+DB). Hence we get
(1) r^2 =r1^2+ r2^2.
(2) Since BE=sqrt(2) *r and
FG^2 =(r2-r1)^2+(r2+r1)^2 =2(r1^2 +r2^2),
we get BE=FG.
(3) Let X and Y be points on BA and BC, respectively.
Since two triangles ABD and DC are similar,
FD/XF=GD/YG, that is, XY // FG.
Now, we see that XY is perpendicular to BE and hence FG is perpendicular to BE.
To Bae deo rak
ReplyDeleteIt is not clear to me how do you get r^2 =r1^2+ r2^2 from r=(AB*BC)/(AB+BC+CA),r1=(AD*BD)/(AB+BD+DA) and r2=(CD*BD)/(BC+CD+DB)
Please explain
To Bae Deok Rak
ReplyDeleteIn (3), what are point X, Y? Without defining their positions, it is impossible to understand the proof.
Please answer my question.
http://img821.imageshack.us/img821/6514/problem25.png
ReplyDeleteHope that my comment below will answer to questions of Peter and Nilton Lapa
Let r, r1, r2 are radii of incircles of triangles ABC, ADB and BDC ( see sketch)
1. Since these triangles are similar so r1= r. (AB/AC) and r2=r.(BC/AC)
We have r1^2+r2^2= r^2.(AB^2+BC^2)/AC^2
But AC^2=AB^2+BC^2 so r1^2+r2^2= r^2
BE^2= 2*r^2 and FG^2 =(r2-r1)^2+(r2+r1)^2 =2(r1^2 +r2^2)= 2.r^2
So BE=FG
2. Let alpha= ∠ (ACB)= ∠ (ABD) and x= ∠ (KGH), y=∠ (KBE)
Since ADB similar to BDC so r1/r2=AB/BC= tan(alpha)
We have tan(x)=(r2-r1)/(r2+r1)= (1-tan(alpha))/(1+tan(alpha))
Tan(y)= tan( m(ABE)-m(ABD))= tan(45-alpha)= (1-tan(alpha))/(1+tan(alpha))
So x=y
Since ∠ (KHG) complement to x so ∠ (KHG) complement to y .
In triangle BHL, ∠ (KHG) complement to ∠ (KBE) so BE perpendicular to FG
I had the idea below, I’d like to know your opinion.
ReplyDeleteLet L on AB, and M on BC, be the tangent points of the incircle of ABC. Let P on AB be the tangent point of the incircle of ABD, and let Q on BC be the tangent point of the incircle of BDC. BLEM is a square with side r, which diagonals LM and BE are congruent and perpendicular. From problem 023, BD=r+r1+r2. As LP=BP-BL, BP=BD-r1, and BD=r+r1+r2 (this is the result of Problem 023), then LP=r2. Similarly, MQ=r1. So, LPF and GQM are congruent, with LF=MG. As LP and QG are paralel, and PF and MQ are paralel too, then LF and MG are paralel. This shows that LMGF is a paralelogram, with LM and FG congruent and paralel. But LM and BE are perpendicular, so BE and FG are congruent and perpendicular.
FG^2 = (r1-r2)^2 + (r1+r2)^2 = 2(r1^2+r2^2) = 2r^2 by referring to my proof for Problem 24
ReplyDeleteBut BE^2 = 2r^2 by considering the square whose diagonal is BE
Hence BE = FG
For the 2nd part of this problem I salute Nilton Lapa's excellent solution above.
Sumith Peiris
Moratuwa
Sri Lanka
FG//EL and FG=EL
ReplyDeleteSee my solution at https://www.facebook.com/photo.php?fbid=10206229971456879&set=a.10205987640598759.1073741831.1492805539&type=3&theater
The first part:
ReplyDeleteFrom the point E draw a line perpendicular onto the line BD. The line meets at the point Q. In problem 23 we have BD= r+r1+r2, therefore :
BQ=BD-r=r1+r2
and from problem 28 we have
EQ= r2-r1
Draw a line from the point G perpendicular onto AC. The line meets at the point N.
Then draw a line from the point F perpendicular onto GN. The line meets at the point M.
Then we have :
FM=r1+r2
and
GM = r2-r1
The triangles FMG and BQE are congruent from SAS and FG=BE
Second part :
We have :
∠GFM =∠EBQ=u
Let the line FG meet the line BD at the point P, then :
∠FPD=∠BPG=90-u
If we extend the line BE to meet FG at a point K and let :
∠BKP=x
than the angles of the triangle of BKP :
180=∠BKP+∠EBQ+∠BPK
180=x+u+90-u
90=x