Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 961.
Sunday, January 19, 2014
Geometry Problem 961. Quadrilateral, Trisection, Sides, Sum of Areas
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Denote (XYZ)= area of XYZ
ReplyDeleteSince AP/AB=AN/AD= 1/3 => (APN)=1/9(ABD)
Note that PNUS is a parallelogram
So (PRN)=(SRU)
So S1= (ANRP)=1/9(ABD)+(SRU)
Similarly S3= (FTGC)=1/9(BCD)+(STU)
And S1+S3= 1/9(ABCD)+ (STUR)
In the same way as above we also have
S2=(BQSE)= 1/9(ABC)+ (SRT) and S4=(MDHU)=1/9(ADC)+(RTU)
And S2+S4=1/9(ABCD)+(STUR)
So S1+S3=S2+S4
Use the same R,S,T,U as in Problem 960.
ReplyDeleteLet S(ABC) denote the area of ABC.
S1 = S(APR) + S(ANR)
= 1/3 [S(ABR) + S(ADR)]
= 1/9 [S(ABE) + 2 S(ABN) + S(ADH) + 2 S(ADP)]
= 1/27 [S(ABC) + 4 S(ABD) + S(ADC)]
= 1/27 [S(ABCD) + 4 S(ABD)]
Similarly,
S2 = 1/27 [S(ABCD) + 4 S(ABC)]
S3 = 1/27 [S(ABCD) + 4 S(BCD)]
S4 = 1/27 [S(ABCD) + 4 S(ACD)]
Thus,
S1 + S3 = 1/27 [2 S(ABCD) + 4 S(ABD) + 4 S(BCD)]
= 1/27 [2 S(ABCD) + 4 S(ABCD)] = 2/9 S(ABCD)
Similarly,
S2 + S4 = 2/9 S(ABCD)
Hence, S1 + S3 = S2 + S4.