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From: luong@tonto.DEC (Van Luong Nguyen  UHO  DTN 264-6560)
Newsgroups: net.math
Subject: Gemometry III puzzle solution
Message-ID: <2613@decwrl.UUCP>
Date: Mon, 16-Jul-84 17:26:51 EDT
Article-I.D.: decwrl.2613
Posted: Mon Jul 16 17:26:51 1984
Date-Received: Tue, 17-Jul-84 04:27:03 EDT
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>Given right triangle ABC (A being the right angle), with point D on AC such
>that BD=CD=1, and AF being the perpendicular from A to BC. If CF=1, find AC.

Solution:	Let angle ACB = x 
--------
Since angle AFC is 90 degrees, we have :

	CF/AC = sin x   therefore, as CF=1:     AC = 1/cos x		(1)

Now, DC=DB, therefore angles DBC=DCB=x, and it follows that:

	angle BDA = 180 - BDC = 180 - (180-2x) 
	      BDA = 2x
Therefore:
	AD/BD = cos 2x 

Since BD=1  and   cos 2x = [2(cosx)(cosx) - 1] ,
	AD = [2(cosx)(cosx) - 1]
and:
	AC = AD + DC = [2(cosx)(cosx) - 1] + 1 = 2(cosx)(cosx)             (2)

Combining (1) and (2) gives:  1/cosx = 2(cosx)(cosx)
or:				cosx = (1/2)**(1/3)
and finally:
		AC = 1/cosx = 2**(1/3)         (third root of 2)
		   = 1.25992....



Van Luong Nguyen,
Digital Equipment Corporation,
Nashua, NH.