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# Find the area....

Triangle ABC is divided into four parts, and the area of each part is as shown in the figure. Find the value of x, where x is the area.

Note by Kshitij Johary
4 years, 1 month ago

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Let the cevians with endpoints $$B$$ and $$C$$ intersect $$AC$$ and $$AB$$ at $$I_\text{swag}$$ and $$I_\text{yolo}$$, respectively. Let $$BI_\text{swag}$$ and $$CI_\text{yolo}$$ intersect at $$I_\text{sa}$$. Let $$area(\triangle AI_\text{sa}I_\text{yolo})$$ be $$A_\text{bad}$$ and $$area(\triangle AI_\text{sa}I_\text{swag})$$ be $$A_\text{toed}$$. Condsider triangles $$\triangle AI_\text{yolo}B$$ and $$\triangle AI_\text{yolo}I_\text{sa}$$. Since they share the same altitude, creary $\frac{A_\text{bad}}{9+A_\text{bad}+A_\text{toed}}=\frac8{8+12}.$ Similarly, consider triangles $$\triangle AI_\text{swag}C$$ and $$\triangle AI_\text{swag}I_\text{sa}$$. Since they share an altitude, creary $\frac{A_\text{bad}}{A_\text{bad}+A_\text{toed}+8}=\frac9{9+21}.$ Solving for $$A_\text{bad}$$ and $$A_\text{toed}$$ gives $$A_\text{bad}=21$$ and $$A_\text{toed}$$, so the answer is $$A_\text{bad}+A_\text{toed}={\huge A_\text{BAD TOED}}=41$$.

EDIT: why is this getting downvoted? It's perfectly legit.

- 4 years, 1 month ago

I think I've learned a little from you. Thank you. :)

- 4 years ago

u nailed it!

- 4 years ago

- 4 years, 1 month ago

yolo sa

EDIT: Ok I'll actually elaborate the most confusing part.

So note that triangles $$\triangle I_\text{yolo}CI_\text{sa}$$ and $$\triangle BI_\text{sa}C$$ have the same height so the ratio of their areas is the same as the ratio of their bases. Thus $\frac{I_\text{yolo}I_\text{sa}}{I_\text{yolo}I_\text{sa}+I_\text{sa}B}=\frac8{8+12}.$ So that's how I got $\frac{A_\text{bad}}{9+A_\text{bad}+A_\text{toed}}=\frac8{8+12}.$

- 4 years, 1 month ago

Okay, let me give an explanation with slightly less confusing variable names. We use the notation $$[X]$$ to denote the area of triangle $$X$$.

Let the point on side $$AB$$ which is hit by the line emanating from $$C$$ be $$K$$. Define $$L$$ on $$AC$$ similarly. Let $$M$$ be the intersection of lines $$CK$$ and $$BL$$. Let the areas of $$\triangle AKM$$ and $$\triangle ALM$$ be $$R$$ and $$S$$ respectively.

Notice that $$\frac{[ALM]}{[LMC]} = \frac{[ABL]}{[BLC]} = \frac{AL}{LC}$$. In terms of $$R$$, $$S$$, and numbers, we get that $\frac{S}8 = \frac{9+R+S}{12+8}, \quad \text{or} \\ \frac{S}{9+R+S} = \frac{8}{12+8}.$ Also, notice that $$\frac{[AKM]}{[KMB]} = \frac{[ACK]}{[CKB]} = \frac{AK}{KB}$$. In terms of $$R$$, $$S$$, and numbers, we get that $\frac{R}{9} = \frac{8+R+S}{9+12}, \quad \text{or} \\ \frac{R}{8+R+S} = \frac{9}{9+12}.$

Finally, we solve. After simplifying and cross-multiplying both equations, we get $$5S = 18+2R+2S$$ and $$7R = 24+3R+3S$$. From the second equation, we get that $$R = 6+\frac34S$$, and substituting in the first equation, $$5S = 18 + 12 + \frac32S + 2S$$. Hence, $$\frac32S = 30$$, or $$S = 20$$. Substituting this back in the first equation gives $$42=2R$$, or $$21=R$$.

Hence, the requested area is $$20+21 = \boxed{41}.$$

- 4 years, 1 month ago

In a family of 3children what is the probability of having at least one boy

- 1 year, 8 months ago

Unless you want me to find the area of X in the diagram by assuming the diagram is perfectly to scale (which is trivial), I'm convinced there is no mathematical way to find the area given the current information. Though I would love to be proved wrong...

- 4 years, 1 month ago

You're wrong.

- 4 years, 1 month ago

Great proof

- 4 years, 1 month ago