SOLUTION: In triangle RGB, point X divides RG according to RX:XG = 3:5, and point Y divides GB according to GY : Y B = 2 : 7. Let C be the intersection of BX and RY . (a) Find a ratio of wh

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Question 1157509: In triangle RGB, point X divides RG according to RX:XG = 3:5, and point Y divides GB according to GY : Y B = 2 : 7. Let C be the intersection of BX and RY .
(a) Find a ratio of whole numbers that is equal to the area ratio CGB : CBR.
(b) Find a ratio of whole numbers that is equal to the area ratio CBR:CRG.
(c) Find a ratio of whole numbers that is equal to the area ratio CGB : CRG.
(d) Find whole numbers m, n, and p so that CGB:CBR:CRG = m:n:p.
(e) The line GC cuts the side BR into two segments. What is the ratio of their lengths?
Answer by KMST(5422) About Me  (Show Source):
You can put this solution on YOUR website!
Let the intersection of GC and BR, be W .
The triangle with lines BX , RY , and GC=GW looks like this:

It is good to have a diagram to keep track of all the smaller triangles formed inside the original triangle.
We have 6 small triangles, like RCX and XCG , with no smaller triangles inside.
We see triangles red%28CGB%29 , green%28CBR%29 , and blue%28CRG%29 , and see that are the union of 2 of the small triangles.
We have other triangles that are the union of 3 of the small triangles, like GRY and YRB.

Each of lines XB , YR , and WG divides the original triangle RBG into 2 triangular parts,
but those lines are not medians of the original triangle.
A median connects the midpoint of one side that we can consider the base of the triangle to the opposite vertex.
If point Y was the midpoint of GB, sides GY and YB would have the same length, triangles GYR and YRB would have
the same base and height, and as a consequence the same area.
Instead, the ratio of length is GY:YB = 2:7 , or GY/YB=2/7.
GY=%282%2F7%29YB --> highlight%28area%28GYR%29=%282%2F7%29area%28YRB%29%29 .
YB also divides GCB into 2 triangles with the same area ratio:
GY=%282%2F7%29YB --> highlight%28area%28GYC%29=%282%2F7%29area%28YCB%29%29 .
Combining the two equations highlighted above, we get
area%28GYR%29-area%28GYC%29=%282%2F7%29area%28YRB%29-%282%2F7%29area%28YCB%29-->area%28blue%28CRG%29%29=%282%2F7%29%28area%28YRB%29-area%28YCB%29%29-->highlight%28area%28blue%28CRG%29%29=%282%2F7%29area%28green%28CBR%29%29%29 .

XB also divides triangles RGB and CRG into parts with known area ratios:
RX%2FXG=3%2F5-->RX=%283%2F5%29XG-->area%28XCR%29=%283%2F5%29area%28CXG%29
RX%2FXG=3%2F5-->RX=%283%2F5%29XG-->area%28RXB%29=%283%2F5%29area%28XGB%29
Combining the two equations highlighted above, we get
area%28RXB%29-area%28XCR%29=%283%2F5%29area%28XGB%29-%283%2F5%29area%28CXG%29-->area%28green%28CBR%29%29=%283%2F5%29%28area%28XGB%29-area%28CXG%29%29-->highlight%28area%28green%28CBR%29%29=%283%2F5%29area%28red%28CGB%29%29%29

(a) Find a ratio of whole numbers that is equal to the area ratio CGB : CBR.
area%28green%28CBR%29%29=%283%2F5%29area%28red%28CGB%29%29-->5%2F3=area%28red%28CGB%29%29%2Farea%28green%28CBR%29%29%29-->area%28red%28CGB%29%29%3Aarea%28green%28CBR%29%29=highlight%285%3A3%29

(b) Find a ratio of whole numbers that is equal to the area ratio CBR:CRG.
area%28blue%28CRG%29%29=%282%2F7%29area%28green%28CBR%29%29-->7%2F2=area%28green%28CBR%29%29%2Farea%28blue%28CRG%29%29-->area%28green%28CBR%29%29%3Aarea%28blue%28CRG%29%29=highlight%287%3A2%29

(c) Find a ratio of whole numbers that is equal to the area ratio CGB : CRG.
Multiplying the ratios found above:
5%2F3=area%28red%28CGB%29%29%2Farea%28green%28CBR%29%29%29 and 7%2F2=area%28green%28CBR%29%29%2Farea%28blue%28CRG%29%29
-->35%2F6=area%28red%28CGB%29%29%2Farea%28blue%28CRG%29%29-->area%28red%28CGB%29%29%3Aarea%28blue%28CRG%29%29=highlight%2835%3A6%29

(d) Find whole numbers m, n, and p so that CGB:CBR:CRG = m:n:p.
7%2F2=7%2A3%2F%282%2A3%29=21%2F6 , so

Combining that with area%28red%28CGB%29%29%3Aarea%28blue%28CRG%29%29=35%3A6 , we get
area%28red%28CGB%29%29%3Aarea%28green%28CBR%29%29%3Aarea%28blue%28CRG%29%29=35%3A21%3A6-->highlight%28m%3An%3Ap=35%3A21%3A6%29

(e) The line GC cuts the side BR into two segments. What is the ratio of their lengths?
Let a be a=area%28blue%28CRG%29%29%2F6-->6a=area%28blue%28CRG%29%29%2F6
From area%28red%28CGB%29%29%3Aarea%28green%28CBR%29%29%3Aarea%28blue%28CRG%29%29=35%3A21%3A6 , we get
--> area%28RBG%29=35a%2B21a%2B66=62a
Let's focus on highlight%28area%28blue%28CRG%29%29=6a%29 and highlight%28area%28blue%28RBG%29%29=62a%29
Line GC cuts side BR at the point we have called W, into segments RW and WB.
Line GC divides triangle RBG into triangle RWG and triangle WBG,
and divides triangle RBC into triangle RWC and triangle WBC.
The ratio of segments (triangle bases) RW and RB, RW%2FRB is the same as the ratio of the triangle areas:
area%28RWG%29%2Farea%28RBG%29=RW%2FRB --> area%28RWG%29=area%28RBG%29%28RW%2FRB%29 --> area%28RWG%29=62a%28RW%2FRB%29 and
area%28RWC%29%2Farea%28RBC%29=RW%2FRB --> area%28RWC%29=area%28RBC%29%28RW%2FRB%29 --> area%28RWC%29=21a%28RW%2FRB%29
area%28RWG%29=area%28RWC%29%2Barea%28RCG%29 --> 62a%28RW%2FRB%29=21a%28RW%2FRB%29%2B6a --> 62a%28RW%2FRB%29-21a%28RW%2FRB%29=6a --> %2862a-21a%29%28RW%2FRB%29=6a --> %2862-21%29cross%28a%29%28RW%2FRB%29=6cross%28a%29 --> 41%28RW%2FRB%29=6 --> RW%2FRB=6%2F41
Then, WB%2FRB=%28RB-RW%29%2FRB=RB%2FRB-RW%2FRB=1-6%2F41=35%2F41 , and RW%2FWB=%286%2F41%29%2F%2835%2F41%29=6%2F35
So, the ratio of BR segments that we are looking for is highlight%28RW%3AWB=6%3A35%29