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

Labels:
area,
quadrilateral,
side,
trisection

Subscribe to:
Post Comments (Atom)

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.