Challenging Geometry Puzzle: Problem 1600. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.
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Let S(BOD) = X and S(COD) = Y
ReplyDelete(10 + 3) / X = 7 / Y = 20 / (X +Y) ....(1) (simple algebraic manipulation)
10 / 7 = 3 / (X + Y) .....(2)
(1) / (2) ; (13 / X ) / (10 /7) = 20 / 3
Therefore X = 273 / 200
Sumith Peiris
Moratuwa
Sri Lanka
Thanks! 🔍 AEC=10/ADC=7 are from another triangle system. Let's use the original measures - your help is gold! 💛"
ReplyDeleteIf S(△OAC)=x, then S(ODE)=10-x, S(△ODC)=7-x
ReplyDeleteBD:DC=S(△AOB):S(△AOC)=S(△BOD):S(△COD)
Solving 13-x:a=S(△BOD):7-1, we get
S(△BOD)=(x-13)(x-7)/x
Similarly, BE:EA={(13-x)(7-x)/x+(7-x)}:a=3:10-x
Solving this, we get x=(221-√12441)/20
Therefore, S(△BOD)=2.1
Area △AEC : Area △ADC = Area △BEO : Area △BOD
ReplyDeleteProof of the above equation
Delete⊿ABC and points AE, D, O are the same as in the problem.
Let Area⊿AEC=P, Area⊿ADC=Q, Area⊿BEO=R, Area⊿BDO=S,
and let Area⊿AOC=X.
BD : DC = P+R-X : X = S : Q-x holds,
X^2-(P+Q+R+S)X+Q(P+R)=0 ---①
BE : EA = S+Q-X : X = R : P-X holds,
X^2-(P+Q+R+S)X+P(S+Q)=0 ---②
From ① and ②, Q(P+R)=P(S+Q)
Therefore, QR=PS
In other words, P : Q = R : S holds
Area of triangle (BOD) = 61/10
ReplyDelete