reseach a formula

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9772 posts

jayme

#1 h Mar 27, 2025, 8:41 AM

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Dear Mathlinkers,

1.ABCD a square
2. m the lengh of AB
3. M a point on the segment CD
4. 1, 2, 3 the incircles of the triangles MAB, AMD, BMC
5. r1, r2, r3, the radius of 1, 2, 3.

Question : is there a formula with r1, r2, r3 and m?

Sincerely
Jean-Louis

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ND_

#2 h Mar 27, 2025, 9:41 AM

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$r_2 + r_3 = 2m - \frac{m^2}{2r_1}$
$r_1=\frac{m^2}{4m-2r_2-2r_3}$

This post has been edited 1 time. Last edited by ND_, Mar 27, 2025, 10:36 AM

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9772 posts

jayme

#3 h Mar 27, 2025, 9:51 AM

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Dear,

do you have a synthetic proof?

Thank
Jean-Louis

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ND_

#4 h Mar 27, 2025, 10:25 AM

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I only have this proof, which is not synthetic. I don't think that it will have a synthetic proof.

$r_1=\frac{Area}{Semi Perimeter}=\frac{m^2}{m+AM+BM}$
$\iff AM+BM=\frac{m^2}{r_1}-m$

$r_2=\frac{P+B-H}{2}=\frac{AD+DM-AM}{2}$
$r_3=\frac{P+B-H}{2}=\frac{BC+CM-BM}{2}$

$r_2 + r_3 = \frac{AD+DM-AM+BC+CM-BM}{2}=\frac{3m-(AM+BM)}{2}$
$=\frac{3m-\left( \frac{m^2}{r_1}-m \right) }{2}=2m - \frac{m^2}{2r_1}$

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jayme

#5 h Mar 27, 2025, 3:47 PM

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Thank you very much...

Jean-Louis

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9772 posts

jayme

#6 h Mar 27, 2025, 3:50 PM

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Dear,

from your nice result, can we deduce this

1/r1 = 1/(m-2r2) + 1/(m-2r3) ?

Sincerely
Jean-Louis

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ND_

#7 h Mar 28, 2025, 4:43 AM

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ND_ wrote:

$r_1=\frac{m^2}{m+AM+BM}, r_2=\frac{AD+DM-AM}{2}, r_3=\frac{BC+CM-BM}{2}$

$\frac{1}{m-2r_1}=\frac{1}{m-(AD+DM-AM)}=\frac{1}{AM-DM}=\frac{AM+DM}{AM^2-DM^2}=\frac{AM+DM}{m^2}$
Similarly, $\frac{1}{m-2r_2}=\frac{BM+CM}{m^2}$

$\Rightarrow \frac{1}{m-2r_1} + \frac{1}{m-2r_2} = \frac{AM+DM}{m^2}+\frac{BM+CM}{m^2}=\frac{m+AM+BM}{m^2}=\frac{1}{r_1}$

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9772 posts

jayme

#8 h Mar 28, 2025, 5:20 AM

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Dear,
thank you again very much...

Can you send me your name in a private message if you want in order to include it in my article ...

Sincerely
Jean-Louis

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