Dear Hyacinthists,

Dear Jean-Pierre, Vladimir,

Thank you for your solutions.

I think there is a mistake in the book's suggestion.

One has to calculate b-c instead.

Then b-c=(r_b-r_c)tan A/2

I didn't know this result and tried to prove it syntetically.

I failed and tried an algebraic approach.

So (B-C)/2=D and tan D=(b-c)/(r+r_a)=(r_b-r_c)/(b+c)

(see Court, for example).

One has

(b-c)/(r_b-r_c)=(r+r_a)/(b+c)=tan A/2

(b+c)/(r+r_a)=(r_b-r_c)/(b-c)=cot A/2

Two nice results.

Best regards,

Luis

________________________________

To:

Hyacinthos@yahoogroups.com
From:

vndubrovsky@...
Date: Tue, 27 May 2008 01:59:30 +0400

Subject: Re: [EMHL] Triangle construction given and

Dear Luis

If you want just to construct the triangle, not necessarily using the sum

b+c, then it suffices to notice that

r_b-r_c=(b-c) cot A/2, so the problem reduces to the construction of a

triangle from A, r (or r_a) and b-c, which was discussed in post 16341 and

later.

Best regards,

Vladimir

2008/5/26 Luís Lopes <

qed_texte@...>

> Dear Hyacinthists,

>

> Source: an old triangle construction german book.

>

> Its suggestion to both problems is to construct b+c.

> I don't figure out how.

>

> I think that here is more of a matter of knowing a

> previous result than to put the data in a suitable position.

>

> Best regards,

> Luis

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