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.

#Geometry #HelpMe! #MathProblem #Math

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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 $20+21 = \boxed{41}.$

Daniel Whatley - 7 years, 7 months ago

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.

Ryan Soedjak - 7 years, 7 months ago

u nailed it!

Jun Das - 7 years, 7 months ago

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

Adam Zaim - 7 years, 7 months ago

Your answer isn't clear "at least to me".

Ilyas Hamo - 7 years, 7 months ago

Bad toed

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 $\frac{A_\text{bad}}{9+A_\text{bad}+A_\text{toed}}=\frac8{8+12}.$

Ryan Soedjak - 7 years, 7 months ago

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

Dikshit Garg - 5 years, 3 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...

Michael Tong - 7 years, 7 months ago

You're wrong.

Ryan Soedjak - 7 years, 7 months ago

Great proof

Matt McNabb - 7 years, 7 months ago

@Matt McNabb – I had fun writing it.

Ryan Soedjak - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$