laura@utcsstat.UUCP (Laura Creighton) (10/15/83)
Laura here. There has been a shuffle at purdue. Hao-Nhein still has an account.
You can mail to me and i will forward to him, if you like, or wait for his
new emergence from a *NEW ACCOUNT*. In the mean time, here is the summary of
solutions to his high school problem.
Laura Creighton, utcsstat!laura
***************** begin forwarded message **********************
I received some solutions to the posted problems and now that
there is no more incoming responses, I decided to post the solution.
Let me make it clear that solving a problem in 5 or 10 or whatever, means
that the total time between reading the problem and completing the solution
is 5 or 10 minutes. If you "see the right gestalt", the problem takes only
3-4 minutes, but "the right gestatlt" is not easy to see.
I have several solutions using trigonometry, and Mark Brader
and Saumya Debray also solved the problem by trig, arriving at a long
trig expression that numerical substitution yields 10 degrees. Here are the
solutions by Spencer Thomas using synthetic geometry and one by Peter
Montgomery using sine rule to arrive at a geometrical answer.
Stuffs in brackets [ ] are mine.
Problem: ABC isoscele (sp?) triangle, B=C=80 degrees, D on AB, AD=BC.
Find angle ACD.
Solution by Spencer Thomas (utah-cs!thomas@UTAH-GR)
---------------------------------------------------
You just "walk" the line segment BC up the triangle, creating isoscele
triangles with their top vertices on alternating sides of the triangles, with
successive top angles of 20, 60, 100, and 140. The last triangle has A as
a vertex, its top vertex is D (the vertex with the 140 degree angle). C and
D are base vertices of an isosceles triangle with a top angle of 160, thus
the angle ACD is 10 degrees.
=Spencer
[Note: The second isosceles triangle is actually an equilateral triangle:
that is how you get C, D as base vertices of an isoscele triangle.]
Solution by Peter Montgomery
-------------------------------------------------
By the law of sines we may assume:
AB = AC = sin(80) (all angles are in degrees)
BC = AD = sin(20)
Then
AD sin(20) AC
_______ = _______ = 2 cos(10) = 2 AC = _______
sin(10) sin(10) sin(150)
[The last step is simpler than I thought when I first read it. It merely follows
from sin(150) = 1/2 ]
Since angle DAC = 20, this proves triangle DAC is similar to one with angles
150 degrees, 20 degrees, 10 degrees (two proportional sides and angle
between them). Hence angle ACD = 10 degrees.
Peter Montgomery
{bli,blix,bmcg,burdvax,cbosgd,csun,hplabs,hughes,ihnp4,ihnss,
netvax,orstcs,parallax,randvax,sdccsu3,sdcnet,sdcsvax,slant45,
sytek,trw-unix,ucla-s,ucla-vax}!sdcrdcf!pmontgom
================================================================================
Hao-Nhien Qui Vu (pur-ee!vu, or pur-ee!norris, or pur-ee!Pucc-H:dlk)
[don't expect me to have a pathname longer than Montgomery's]