All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
Congruent Incircles (Posted on 2017-01-01) Difficulty: 4 of 5
Let ABC be a 3-4-5 triangle with right angle C. Let D be a point on the hypotenuse. CD then partitions ABC into ACD and BCD.

Where is D located if ACD and BCD have congruent incircles?

No Solution Yet Submitted by Brian Smith    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution numerical computer solution | Comment 4 of 6 |
The radius of an incircle of a triangle is given by

r = A/s

where A is the area of the triangle and s is the semi-perimeter.

Heron's formula gives the area of the triangle:

A = sqrt(s(s-a)(s-b)(s-c))

when the area is divided by s to get the inradius, you get

r = sqrt((s-a)(s-b)(s-c)/s)

In order for the incircles to be congruent, the results of this formula must be the same for each side (each subtriangle).

The program assumes A is the angle at the end of the 3-unit leg, opposite the 4-unit leg.

DefDbl A-Z
Dim crlf$


Private Sub Form_Load()
 Form1.Visible = True
 
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)
 
 angle = Atn(4 / 3)
 ' angle at A, where hypotenuse meets side of length 3
 For ad = 2.2898979 To 2.289898 Step 0.000000001
   DoEvents
   cd = Sqr(ad * ad + 9 - 6 * ad * Cos(angle))
   peri1 = 3 + ad + cd
   peri2 = 4 + 5 - ad + cd
   s1 = peri1 / 2: s2 = peri2 / 2
   r1 = Sqr((s1 - 3) * (s1 - ad) * (s1 - cd) / s1)
   r2 = Sqr((s2 - 4) * (s2 - (5 - ad)) * (s2 - cd) / s2)
   Text1.Text = Text1.Text & mform(ad, "0.000000000")
   Text1.Text = Text1.Text & mform(r1, "  0.00000000")
   Text1.Text = Text1.Text & mform(r2, "  0.00000000")
   If r2 > r1 Then Text1.Text = Text1.Text & " *"
   Text1.Text = Text1.Text & crlf
 Next
  
  
 Text1.Text = Text1.Text & crlf & " done"
  
End Sub

Function mform$(x, t$)
  a$ = Format$(x, t$)
  If Len(a$) < Len(t$) Then a$ = Space$(Len(t$) - Len(a$)) & a$
  mform$ = a$
End Function

which has already had its limits and step size for AD narrowed down, finds AD needing to be between 2.289897948 and 2.289897949, resulting in both inradii being 0.71010205.

Relevant section of output:

     AD              In-radii   
2.289897944  0.71010205  0.71010205 *
2.289897945  0.71010205  0.71010205 *
2.289897946  0.71010205  0.71010205 *
2.289897947  0.71010205  0.71010205 *
2.289897948  0.71010205  0.71010205 *
2.289897949  0.71010205  0.71010205
2.289897950  0.71010205  0.71010205
2.289897951  0.71010205  0.71010205
2.289897952  0.71010205  0.71010205
2.289897953  0.71010205  0.71010205

radius numbers differ beyond shown digits, but an asterisk marks where r2 > r1.

As this uses the opposite acute angle as vertex A it is 5 minus Bractal's value for AD. So the two answers are in agreement.

  Posted by Charlie on 2017-01-01 12:43:06
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (7)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information