See complete Problem 100
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry, Tutoring
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Geometry Problem 100
Subscribe to:
Post Comments (Atom)
This follows quite trivially from the fact that AC²+BC²+2CD²=AD²+BD²=AB².
ReplyDeletename x radius for AC, y for CB, z for CD, R for AB
ReplyDeletename Sa area above 1 & 2, Sb left of 1, Sc right of 2
S1 + Sb = x²/2 (1)
S2 + Sc = y²/2 (2)
S1 + S2 + S3 + Sa = x∙y (3)
( from euklid theorem (2z)²=2x∙2y => z² = x∙y )
S1 + S2 + S3 + S4 + S5 + Sa + Sb + Sc = R²/2 + S3
(S1+Sb)+(S2+Sc)+(S1+S2+S3+Sa)-S1-S2+S4+S5 = R²/2 + S3
substitute (1), (2), (3)
(x²/2 + y²/2 + x∙y) - S1 - S2 + S4 + S5 = R²/2 + S3
R²/2 - S1 - S2 + S4 + S5 = R²/2 + S3
(R=x+y, R²=x²+2xy+y², R²/2 = x²/2 + xy + y²/2)
S4 + S5 = S1 + S2 + S3
Let S(AB), S(AC), S(BC) and S(CD) be the areas of the semicircles with diameters AB, AC, BC and CD, respectively.
ReplyDeleteWe have S(AB) = 2S(CD) – S3 + S4 + S5 +
+ S(AC) – S1 + S(BC) – S2, so
S1 + S2 + S3 = S4 + S5 + S(AC) + S(BC) +
+ 2S(CD) – S(AB).
Thus, to prove that S1 + S2 + S3 = S4 + S5, it’s enough to prove that
S(AB) = S(AC) + S(BC) + 2S(CD).
We have S(AC) + S(BC) + 2S(CD) =
= (pi/2)[AC^2/4 + BC^2/4 + 2.CD^2/4] =
= (pi/2)[AC^2/4 + CD^2/4 + BC^2/4 + CD*2/4] =
= (pi/2)[(AD^2/4 + BD^2/4] = (pi/2)(AB^2/4) =
= S(AB).
Q.E.D.
let the midpoint of AB is O, connect OD
ReplyDeletelet "L","M","N"and"R" be the names of circles having the diameter AB,CB,AC and CD respectively
say CB/2=x & AC/2=y
so , radius of "L"=2(x+Y)/2=x+y
apply Pythagoras theorem in triangle DCO
so the radius of "R"=2√xy/2 = √xy
Area of L=1/2∏(x+y)^2
area of M=1/2∏x^2
area of N=1/2∏y^2
L-M-N = ∏xy
area of R= ∏xy
so area of R= area of L-M-N
R- (R-(S1+S2+S3)) = (L-M-N)- (R-(S1+S2+S3))
so,S1+S2+S3=S4+S5