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
6 years ago

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

Let the point on side ABAB which is hit by the line emanating from CC be KK. Define LL on ACAC similarly. Let MM be the intersection of lines CKCK and BLBL. Let the areas of AKM\triangle AKM and ALM\triangle ALM be RR and SS respectively.

Notice that [ALM][LMC]=[ABL][BLC]=ALLC\frac{[ALM]}{[LMC]} = \frac{[ABL]}{[BLC]} = \frac{AL}{LC}. In terms of RR, SS, and numbers, we get that S8=9+R+S12+8,orS9+R+S=812+8.\frac{S}8 = \frac{9+R+S}{12+8}, \quad \text{or} \\ \frac{S}{9+R+S} = \frac{8}{12+8}. Also, notice that [AKM][KMB]=[ACK][CKB]=AKKB\frac{[AKM]}{[KMB]} = \frac{[ACK]}{[CKB]} = \frac{AK}{KB}. In terms of RR, SS, and numbers, we get that R9=8+R+S9+12,orR8+R+S=99+12.\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+2S5S = 18+2R+2S and 7R=24+3R+3S7R = 24+3R+3S. From the second equation, we get that R=6+34SR = 6+\frac34S, and substituting in the first equation, 5S=18+12+32S+2S5S = 18 + 12 + \frac32S + 2S. Hence, 32S=30\frac32S = 30, or S=20S = 20. Substituting this back in the first equation gives 42=2R42=2R, or 21=R21=R.

Hence, the requested area is 20+21=41.20+21 = \boxed{41}.

Daniel Whatley - 5 years, 12 months ago

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Let the cevians with endpoints BB and CC intersect ACAC and ABAB at IswagI_\text{swag} and IyoloI_\text{yolo}, respectively. Let BIswagBI_\text{swag} and CIyoloCI_\text{yolo} intersect at IsaI_\text{sa}. Let area(AIsaIyolo)area(\triangle AI_\text{sa}I_\text{yolo}) be AbadA_\text{bad} and area(AIsaIswag)area(\triangle AI_\text{sa}I_\text{swag}) be AtoedA_\text{toed}. Condsider triangles AIyoloB\triangle AI_\text{yolo}B and AIyoloIsa\triangle AI_\text{yolo}I_\text{sa}. Since they share the same altitude, creary Abad9+Abad+Atoed=88+12.\frac{A_\text{bad}}{9+A_\text{bad}+A_\text{toed}}=\frac8{8+12}. Similarly, consider triangles AIswagC\triangle AI_\text{swag}C and AIswagIsa\triangle AI_\text{swag}I_\text{sa}. Since they share an altitude, creary AbadAbad+Atoed+8=99+21.\frac{A_\text{bad}}{A_\text{bad}+A_\text{toed}+8}=\frac9{9+21}. Solving for AbadA_\text{bad} and AtoedA_\text{toed} gives Abad=21A_\text{bad}=21 and AtoedA_\text{toed}, so the answer is Abad+Atoed=ABAD TOED=41A_\text{bad}+A_\text{toed}={\huge A_\text{BAD TOED}}=41.

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

Ryan Soedjak - 6 years ago

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u nailed it!

Jun Das - 5 years, 11 months ago

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

Adam Zaim - 5 years, 11 months ago

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

Ilyas Hamo - 6 years ago

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Bad toed

yolo sa

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

So note that triangles IyoloCIsa\triangle I_\text{yolo}CI_\text{sa} and BIsaC\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 IyoloIsaIyoloIsa+IsaB=88+12.\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 Abad9+Abad+Atoed=88+12.\frac{A_\text{bad}}{9+A_\text{bad}+A_\text{toed}}=\frac8{8+12}.

Ryan Soedjak - 6 years ago

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

Dikshit Garg - 3 years, 7 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 - 6 years ago

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You're wrong.

Ryan Soedjak - 6 years ago

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Great proof

Matt McNabb - 5 years, 12 months ago

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@Matt McNabb I had fun writing it.

Ryan Soedjak - 5 years, 12 months ago

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