| Newsgroups: net.math,net.puzzle
Path: decwrl!decvax!ittvax!dcdwest!sdcsvax!akgua!mcnc!ecsvax!pizer
Subject: Solution to Geometry III
Posted: Sun Jul 15 07:24:22 1984
References: <ecsvax.2915>
Here's the solution to the third problem:
#3
>>First start with right triangle ABC, where A is rt. and where AC is the
>>longer leg. Then draw point D on side AC, such that DB=DC. Finally, draw
>>point F on BC such that AF is perpendicular to BC, and such that CF=CD. Given:
>>BD=DC=FC=1 and that AB is prpndclr to AC and that AF is prpndclr to BC, FIND:
>>AC. Good luck, leave answer in radical form.
My solution to this problem was simple, but used a lot of algebra, it goes as
follows:
Let X=AC
AF = SQRT(AC^2 - FC^2) AF = SQRT(X^2 - 1)
BA = SQRT(BD^2 - AD^2) BA = SQRT(1 - (X-1)^2)
BF = SQRT(BA^2 - AF^2) BF = SQRT(1 - (X-1)^2 - (X^2 - 1))
BC = BF + FC BC = SQRT(1 - (X-1)^2 - (X^2 - 1)) + 1
BC^2 = BA^2 + AC^2
--BC^2-- --BA^2-- --AC^2--
(SQRT(1 - (X-1)^2 - (X^2 - 1)) + 1)^2 = 1 - (X-1)^2 + X^2
(SQRT(1 - (X^2 - 2X + 1) - X^2 + 1) + 1)^2 = 1 - (X^2 - 2X + 1) + X^2
(SQRT(-2X^2 + 2X + 1) + 1)^2 = 2X
-2X^2 + 2X + 1 + 1 + 2SQRT(-2X^2 + 2X + 1) = 2X
2SQRT(-2X^2 + 2X + 1) = 2X^2 - 2
SQRT(-2X^2 + 2X + 1) = X^2 - 1
-2X^2 + 2X + 1 = X^4 - 2X^2 + 1
X^4 - 2X = 0
X^3 - 2 = 0 (X cannot be 0)
X^3 = 2
X = 2^1/3
. AC equals 2^(1/3)
. . Q.E.D.
Another good answer was sent to me by Kamal Abdali (allegra!tekchips!abdali)
and goes as follows:
>Let the length AD be denoted by x, and the angle ACB by theta. From the
>right triangle ACF, we get cos(theta) = CF/AC, i.e.
> cos(theta) = 1/(1+x). (1)
>The angle ADB equals the sum of angles DBC and DCB. So angle ADB is
>2*theta. From right triangle ADB, we get cos(2*theta) = AD/DB, i.e,
> cos(2*theta) = x. (2)
>Write c for cos(theta), hence 2*c^2-1 for cos(2*theta). Then (1) and
>(2) become:
> c = 1/(1+x), that is, x = 1/c - 1,
> 2*c^2 - 1 = x.
>Combining, we get 2*c^2 = 1/c, i.e, c = 1 / 2^(1/3). Now AC=AD+DC=1+x=1/c
>by (1). Hence, the answer is AC = cuberoot(2).
Several other people sent me different methods of solving the problem, however,
much to my dismay, I don't have room to list their solutions. Thank you to
those people for taking the time to solve the problems and send me their ideas.
Billy Pizer
({decvax,akgua,ihnp4,burl}!mcnc!ecsvax!pizer)
-- "What holds things together?"
"Neutrons and protons are held together by velcro." --
|
| Newsgroups: net.math
Path: decwrl!dec-rhea!dec-tonto!luong
Subject: Gemometry III puzzle solution
Posted: Mon Jul 16 14:26:51 1984
>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.
|