Proposed Problem
Try to use elementary geometry (Euclid's Elements)
See complete Problem 19 at:
gogeometry.com/problem/problem019.htm
Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Saturday, February 21, 2009
Problem 19: Isosceles Right Triangle, Excenter, Perpendicular, Congruence
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Let B:(0,0), A:(0,a) and C:(a,0)
ReplyDeleteArea ABC=a^2/2, AC=V2*a where V=square root
Hence exradius =(a^2/2)/(2a -v2a) = a/(2-v2)
Hence AD^2 = (a/(2-v2))^2 + (a/v2)^2
which gives AD ~ 1.8477590650*a
while BE=a*cos(ABE)=acos(22.5)~ 0.923879532*a
2BE = 2*0.923879532*a = 1.847759065*a
Hence AD = 2BE
Ajit: ajitathle@gmail.com
from triangle BEA BE=AE*tan67.5, from triangle BED BE=(AE+AD)*tan22.5,then
ReplyDeleteAE=BEcot67.5 and AE=BE*cot22.5-AD. By using last two relations we get BE*cot67.5=BE*cot22.5-AD, then BE(tan67.5-tan22.5)=AD, BE(sin(67.5-22.5)\(cos67.5*cos22.5))=AD , BE(sin45\(sin22.5*cos22.5))=AD then AD=2BE
Shahlar Maharramov, Yasar University ,Izmir,Turkey and Cybernetics Institute Azerbaijan
http://img534.imageshack.us/img534/2876/problem19.png
ReplyDeleteBE cut DF at G.
Note that BFDH is a square >> BF=DF ( see picture)
Triangle BFG congruence to Tri. DFA (Case SAS)
So AD=BG
In Triangle GDB , ED is an angle bisector and ED is an altitude so BGD is isosceles
So BG=2BE=AD
Peter Tran
Complete the square BFDG and let DF & BE meet at H. Then Tr.s HFB & AFD are congruent SAA. So the 2 hypotenuses are equal. BE = EH since DE is both the altitude and the angle bisector
ReplyDeleteSo 2BE = AD
Sumith Peiris
Moratuwa
Sri Lanka
Let F be the tangent point with AB and G the tangent point with AC and s be the length of AB.
ReplyDelete1. AG is sqrt(2)/2 * S.
2. tr ADF is congruent to ADH (2 sides and they are both right triangles) so AF = AG = sqrt(2)/2 * S.
3. DF and BF are both the radius length of the circle and = (1 + sqrt(2)/2) S
4. Using the Pythagorean theorem AD is therefore sqrt(2 + sqrt(2))
5. Since ABE is similar to ADF we can just use ratios to find BE = (1 / sqrt(2 + sqrt(2)) * ((2 + sqrt(2))/2) = sqrt(2 + sqrt(2))/2 = AD /2
See original solution on this video
ReplyDelete