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 Congruent Incircles (Posted on 2017-01-01)
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)

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 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\$

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
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:
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  0.710102052.289897950  0.71010205  0.710102052.289897951  0.71010205  0.710102052.289897952  0.71010205  0.710102052.289897953  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

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