My Math Forum Split area

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 July 29th, 2012, 10:27 AM #1 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Split area An "all integer" example: isosceles triangle; AB = AC = 442 ; BC = 84. D is on AB, 104 from B; E is on AC, 153 from C. Line DE = 77. Triangle ADE's area = half of triangle ABC's area. Code:  A 289 338 E 77 D 153 104 B 84 C As general case: given is AB (or AC), BC and BD; make 'em : AB=AC=a, BC = b and BD = d Calculate DE in terms of a,b,d Just for something to do
 July 31st, 2012, 01:01 PM #2 Senior Member   Joined: Jul 2011 Posts: 118 Thanks: 0 Re: Split area $A_{\small ADE}=\frac{1}{2}|AD||AE|\sin\angle BAC=\frac{1}{2}A_{\small ABC}=\frac{1}{4}|AB||AC|\sin\angle BAC |AD||AE|=\frac{1}{2}|AB||AC| \big(|AB|-|BD|\big)|AE|=\frac{1}{2}|AB|^2 |AE|=\frac{|AB|^2}{2\big(|AB|-|BD|\big)}=\frac{a^2}{2(a-d)} \text{cosine formula:} |BC|^2=|AB|^2+|AC|^2-2|AB||AC|\cos\angle BAC \cos\angle BAC=\frac{|AB|^2+|AC|^2-|BC|^2}{2|AB||AC|}=\frac{2a^2-b^2}{2a^2} |DE|^2=|AD|^2+|AE|^2-2|AD||AE|\cos\angle BAC |DE|=\sqrt{(a-d)^2+$$\frac{a^2}{2(a-d)}$$^2-2$$\frac{1}{2}a^2$$$$\frac{2a^2-b^2}{2a^2}$$}= =\sqrt{(a-d)^2+\frac{a^4}{4(a-d)^2}-a^2+\frac{1}{2}b^2}$
 July 31st, 2012, 01:18 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Re: Split area Looksee good! Mine: AD = f = a - d (given) angleBAC = u = 2ASIN[b/(2a)] (given) CE = e = [4afSIN(u) - bSQRT(4a^2 - b^2)] / [4fSIN(u)] DE = SQRT[(a - e)^2 + f^2 - 2f(a - e)COS(u)]

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