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.

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TopNewestLet 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 · 3 years, 10 months ago

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– Adam Zaim · 3 years, 9 months ago

I think I've learned a little from you. Thank you. :)Log in to reply

– Jun Das · 3 years, 9 months ago

u nailed it!Log in to reply

– Ilyas Hamo · 3 years, 10 months ago

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

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 · 3 years, 10 months ago

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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 · 3 years, 10 months ago

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In a family of 3children what is the probability of having at least one boy – Dikshit Garg · 1 year, 5 months ago

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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 · 3 years, 10 months ago

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– Ryan Soedjak · 3 years, 10 months ago

You're wrong.Log in to reply

– Matt McNabb · 3 years, 10 months ago

Great proofLog in to reply

– Ryan Soedjak · 3 years, 10 months ago

I had fun writing it.Log in to reply