| Newsgroups: net.math,net.puzzle
Path: decwrl!decvax!mcnc!unc!ulysses!mhuxl!houxm!ihnp4!gargoyle!oddjob!matt
Subject: Re: Geometry Puzzle I SOLUTION
Posted: Mon Jul 9 13:18:52 1984
Given information:
BAC = 40, ABE = ADE = 10, AB = AD, B != D.
(1) Exterior angle thm ==> AEB = BAC - ABE = 40 - 10 = 30
(2) sin AEB sin ABE
------- = ------- (law of sines)
AB AE
sin ADE
= ------- (given ABE=ADE)
AE
sin AED
= ------- (law of sines)
AD
sin AED
= ------- (given AD=AB)
AB
Therefore sin AEB = sin AED. But B != D ==> AEB != AED,
so AED = 180 - AEB = 180 - 30 = 150
(3) DAE = 180 - ADE - AED = 180 - 10 - 150 = 20.
___________________________________________________________
Matt University ARPA: [email protected]
Crawford of Chicago UUCP: ihnp4!oddjob!matt
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| Newsgroups: net.math,net.puzzle
Path: decwrl!decvax!ittvax!dcdwest!sdcsvax!akgua!mcnc!ecsvax!pizer
Subject: Solution to Geometry I
Posted: Sun Jul 15 07:17:36 1984
References: <ecsvax.2861>,<oddjob.326>
#1
>>It is a semicircle with diameter CAF, with C & F being on the semicircle
>>and A being the center. Point B is a point on the semicircle such that
>>angle BAC is 40 degrees. Point E is a point on segment AF such that angle
>>ABE is 10 degrees. Finally, point D is a point on the semicircle, and not
>>point B such that angle ADE is also 10 degrees. Find angle DAF.
My solution to this problem is to rotate triangle AED around the center, point
A, until it overlaps triangle AEB. Then, lable point E on triangle AED point
E', so there will be no confusion. It becomes apparent that since AE' and AE
are congruent, triangle AEE' is isosceles, with angle E and E' being equal.
Since you know angle BAE is 140 degrees (180-40), and that angle ABE is 10
degrees, angle AEB must be 30, along with angle AE'E. That leaves angle E'AE
to be 180-30*2, or 120, thus making angle BAE' (also angle DAE', since D
overlaps B) 20 degrees. Angle DAE', or just plain DAE when it is rotated back,
is what we're looking for.
Matt Crawford also posted a solution <oddjob.326 (if it is still around)>. He
solved it using the law of sines to prove sin(AEB)=sin(AED), and since we are
told point B is distinct from point D, angle AED must be 180 - AEB, or 150,
making angle DAF 180-150-10, or 20 degrees.
Another solution to this problem, and unfortunately I erased the letter
accidently and can't remember his name, involves extending the semicircle to a
full circle, and then extending BE till it hits the circle, labeled point G.
Then, it becomes apparent that angle GAF is the reflection of angle DAF, and
can then be solved with little trouble. I apologize to the author of this
letter for not giving credit to him.
Billy Pizer
({decvax,akgua,ihnp4,burl}!mcnc!ecsvax!pizer)
-- "Why do the days get longer in the summer?"
"Because heat makes things expand." --
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