A math olympiad problem that looks complicated, but it's easy to solve

[@jessewallis6589]

I used a=100/b; c=200/b; then substituted. Ended up with the right answer.

[@tyfntnc3l]

a=100/b =》ca=c(100/b)=300 =》c=3b =》bc=b(3b)=200 =》3b²=200 =》b=10√2/√3 =》c=3(10√2/√3) =》a=100/10√2/√3 =》a+b+c=(10√3/√2)+(40√2/√3) =》=44,907

[@mashoodbilal7057]

Yes exactly. Me too. 😅

[@rohanclark7020]

Do we have to present our answer in serd form? If no, then.. ab=100 2ab=200 2ab=bc ;eliminate b c=2a ca=300 ;sub c 2aa=300 aa=300/2 aa=150 a=square root of 150 a=(12.24744871) Now find others ab=100 ; Sub for a (square root of 150) x b =100 b = 100/(square root of 150) b=(8.164965809) find c ca=300 ; Sub a c=300/(square root of 150) c=(24.49489743) adding a b and c together does equal 55(sqroot.6)/3 or (44.90731195)

[@СветланаГанеева-х1ж]

Я тоже так решила. Намного легче

[@Краснаяборода-и9у]

100+200+300=600 600:2 (каждая буква встречается два раза)=300(общее число) 300-100(ab)=200(c) 300-200(bc)=100(a) 300-300(ca)=0(b) Так конечно не легче...

[@srmpenedo]

Its easier to calculate a, b and c and then sum them up

[@marknieuweboer8099]

Yup, that took me a minute iso 7 minutes. Also I didn't forgetvthe negative solution.

[@haroonshafi6348]

Yep just 2min

[@vijjwalgupta]

​@@marknieuweboer80997 minutes? it took me around 4 mins

[@vijjwalgupta]

​@@Teronix100i used ur method and ended up getting stucked in ab+bc = ba+bc loop. ur solution is literally incorrect

[@vijjwalgupta]

@@Teronix100 thats exactly what i did before reading ur previous comment or even opening the video after reading the thumbnail

[@davidexel]

Multiply all the equations at once. Then you get (abc)^2 = 6000000 Thus abc = 1000 Sqrt(6) Dividing abc/bc = a = 5 Sqrt(6) Dividing abc/ac = b = 10/3 Sqrt(6) Dividing abc/ab = c = 10 Sqrt(6) Just add a+b+c = 55/3 Sqrt(6)

[@rainerzufall42]

... and then add the second solution a+b+c = - 55/3 Sqrt(6)...

[@МихаилКоган-ч9н]

2ab=bc => c=2a => 2a^2=300 => a=+-5sqrt(6), b=+-10sqrt(6)/3, c=+-10sqrt(6) a+b+c=+-55sqrt(6)/3

[@Sci-Marvels]

Nice work on solving the equation! The steps you've outlined are clear

[@jgarc201]

Sure. This problem has two solutions, one positive and one negative.

[@PlasteredDragon]

You have the same numerical solution that I have, but you've been able to simplify the expression better than I did. I compute a = ±5√6, b = ±20/√6, and c = ±10√6, giving a+b+c = ±110/√6. I have checked to confirm that 110/√6 is indeed equal to (55√6)/3, so we both have the correct value, but clearly I'm forgetting something somewhere about how to simply square roots? Can you tell me what I am missing? My steps were: 1. c in terms of a: ca = 300 --> c = 300/a 2. b in terms of a: bc = 200 --> b = 200/c --> b = 200/(300/a) --> b = 200a/300 --> b = 2a/3 3. solve for a: ab = 100 --> a(2a/3) = 100 --> (2a^2)/3 = 100 --> 2a^2 = 300 --> a^2 = 150 --> a = √150 = (√25)(√6) = 5√6 4. solve for b: ab = 100 --> b = 100/a --> b = 100 / 5√6 = 20/√6 5. solve for c: bc = 200 --> c = 200/b --> c = 200 / (20/√6) --> c = (200√6)/20 --> c = 10√6 6. sum a+b+c = 5√6 + 20/√6 + 10√6 = 15√6 + 20/√6 = 90/√6 + 20/√6 = 110/√6

[@PlasteredDragon]

Nevermind, I figured it out, somewhere along the way you multiply by (√6 / √6) which = 1. You can always multiply by 1, so if I do this: 110/√6 * (√6 / √6) = (110√6)/6 = (55√6)/3. Phew!

[@jgarc201]

@@PlasteredDragon Yes, you have to rationalize, as you have done

[@rainerzufall42]

The negative solution is missing. Be careful!

[@9허공]

simple solution multiplying 3 equations, (abc)^2 = 6*10^6 abc = +-sqrt(6)*10^3 => a = +-sqrt(6)/2*10, b = +-sqrt(6)/3*10, c = +-sqrt(6)/1*10, a+b+c = +-(sqrt(6)*11/6*10)

[@Sci-Marvels]

Great use of multiplication to simplify the problem! Your solution is clear and concise

[@meow0112]

I also came to comment this. Multiplying the equations makes it a lot easier in this case.

[@theotimegalindo9913]

@@Sci-Marvelsit’s missing negative solutions

[@Urgleflogue]

There's no way this is math olympiad problem, I almost solved it in my head.

[@TumeloPercy-z3s]

😂😂😂😂😂😂😂You are lying voetsek your bullshit

[@aisawaloki1571]

I have an simpler way to first find values of a b and c For simplicity assume a, b, and c are 1/10 of their original value first, thus multiple of any 2 of them becomes 1/100 ab=1 bc=2 ca=3 ab*bc*ca=(abc)^2==6 abc=sqrt6 Since sqrt can be negative thus a,b,and c can also be either all positive or all negative, but the calculation are same, thus for easy sake I just demonstrate the positive one. c=abc/ab=sqrt6/1=6/sqrt6 Now convert a,b,and c back to their original value, multiply by 10 each, and can be theorized a=30/sqrt6 b=20/sqrt6 c=60/sqrt6 a+b+c=(60+30+20)/sqrt6=110/sqrt6 Different with the answer in this video? Wait, please change the denominator to 3 from sqrt6 110/sqrt6=(110*sqrt6)/6=(55*sqrt6)/3 Just got the same answer❤ Yes, a, b, and c can be negative so can the answer.

[@rainerzufall42]

So you admit that "the same answer" is not the correct answer? Then why are you so shy to write it down explicitly? The author of the video didn't have the correct answer, and he should know!

[@shawngayner1392]

I had a geometric solution. Draw a right trapezoid with the top being a, the side being b, and the bottom being c. The diagonal is the hypotenuse of a right triangle with sides b and (c-a) which fortunately for us is just a since c = 2a. This triangle also has an area of 50 from (200 - 100)/2. Using the formula for the area of a triangle, 1/2 b*a = 50, and since a = (3/2)b, we can substitute and solve, b = +/- sqrt(200/3). Since a=(3/2)b and c=3b, a+b+c is 5.5b. Take out the sqrt(100) to get +/-55*sqrt2/sqrt3 and simplify. I enjoyed this very much, thank you for the content!

[@Sci-Marvels]

That's a brilliant geometric approach! Visualizing the problem with a right trapezoid and using the properties of the right triangle really adds a new dimension to the solution. I love how you tied it all together with the area calculations and substitutions. Your method is both elegant and insightful. I'm glad you enjoyed the content, and thank you for sharing your unique perspective!

[@erikthenorviking8251]

I am not brilliant at maths, and I am really glad someone envisaged it the same way as me!!

[@cyruschang1904]

1) ab = 100 2) bc = 200 3) ca = 300 1) x 2) x 3) (abc)^2 = 6000000 4) abc = ✓6 x 1000 4) ÷ 1) c = 10✓6 4) ÷ 2) a = 5✓6 4) ÷ 3) b = (10✓6)/3 a + b + c = (18 + 1/3)✓6

[@larryrobx]

Same calc approach as a few other solvers here -- but, prize for the most concise sol'n stmt!

[@rainerzufall42]

Another one with a good approach, but still losing the second solution a + b + c = - (18 + 1/3) ✓6 4) abc = +/- ✓6 x 1000

[@cyruschang1904]

@@rainerzufall42 Yes. You are correct. Thank you!

[@ThachHau1133]

*a/c=1/2=>c=2a* *ac=a×2a=2a^2=300* *a=√150=√(6×5^2) = 5√6*

[@XfeeXg]

ab=100 (*); bc=200(**) ; ca=300(***). Nếu a,b,c khác 0 thì bc=2ab Chia 2 vế cho b ta được c=2a Thay kết quả trên vào(***),ta có: 2a*2=300 a*2=150 a1=+ 5×6^2 ;b1=10×6^2/3; c1=10×6^2 a2= -5×6^2; b2= -10×6^2 /3 ; c2= - 10×6^2.

[@darkmodex0]

I solved by substitution. Took much less time. b=100/a c=2a a=√150=5√6 b=(100/(5√6))=((10√6)/3) c=10√6 ((15√6)/3)+((10√6)/3)+((30√6/3)= ((55√6)/3)

[@rainerzufall42]

Also: a+b+c = - ((55√6)/3) as the second solution...

[@ДенисХалин-г3ь]

я решил задачу так-же как вы, это намного проще) LIKE)))

[@rajveersingh95]

Let, abc = k c = k/100 a = k/200 b = k/300 a + b + c = k(1/100 + 1/200 + 1/300) Also, multiply the orginial 3 eq to get k^2 = 100*200*300 Substitute both the values of k to get ans.

[@richardfarrer5616]

Divide first equation by second to get a/c = 1/2. Similarly b/c = 1/3. Substitute in any formula to get c^2 = 600 so c = 10sqrt(6). Substitute into final sum to get c/2 + c/3 + c = 11c/6 = 110sqrt(6)/6 = 55sqrt(6)/3.

[@ManojkantSamal]

ab/bc=100/200=1/2 a/c=1/2 C=2a...... Eqn1 bc/ca=200/300 b/a=2/3 3b=2a b=2a/3........eqn2 Now ab=100 a×(2a/3)=100 2a^2=300 a^2=150 a=5.*6 (*=Squre root ) b=2a/3=2(5.*6)/3=10.*6/3 C=2a=2(5.*6)=10.*6

[@martinzavalaleon8856]

(a×b×c)^2 = 1×2×3×10^6, Entonces a×b×c = sqrt(6)×10^3 o a×b×c = - sqrt(6)×10^3. c = sqrt(6)×10 b = (sqrt(6)/3)×10 a = (sqrt(6)/2)×10 a + b + c = sqrt(6)×10×(1 + 1/2 + 1/3) = sqrt(6)×10×(11/6) = (55/3)sqrt(6)×10 La otra solución es con los correspondientes negativos.

[@shadeblackwolf1508]

Divide line 2 by line 1, and we get c/a = 2. Multiplying this result with line 3 gives cxc=600, so c=10√6. Therefore a=300/10√6=30/√6=5√6. Then B=100/5√6=20/√6. So A+B+C=5√6+10√6+20/√6=15√6+20/√6. We can rearrange this by multiplying the latter term by 1: (20/√6)×(√6/√6)=20/6×√6. Compacting this further we can say that the sum is 110/6×√6=55/3×√6, or we can go to (18+1/3)√6

[@Sci-Marvels]

Great method! Dividing line 2 by line 1 to get c/a = 2 and then using that to find c = 10sqrt(6) simplifies the process. From there, solving for a and b and adding them up to get 55/3 * sqrt(6) or (18 + 1/3)sqrt(6) is a clean and efficient way to reach the solution.

[@jharvick]

This is a very long way to solve this. Here's a faster solution: multiply all three equations together, you get (abc)^2 = 6,000,000, or, abc=1000rad(6) Divide abc = 1000rad(6) by each of the equations to get each value. Two variables cancel out each time, leaving you with one left. c= 10rad6 a=5rad6 b=10rad(6)/3 c= 30rad(6)/3 a=15rad(6)/3 b=10rad(6)/3 Add them up, 55rad(6)/3

[@darthcalzone6653]

b=100/a c=2a These came from simply rearranging the original formulas a^2=150 I then substituted my value for c into the last equation to give: a=5sqrt(6) #or 15sqrt(6)/3, this becomes important later b=100/5sqrt(6) I substitute this value for a into my new equation b=100/a to get: b=10sqrt(6)/3 I then use my equation c=2a to get: c=10sqrt(6) #or 30sqrt(6)/3 I made them all over 3 to get a common denominator, then just added them up, getting: 15sqrt(6)/3 + 10sqrt(6)/3 + 30sqrt(6)/3 = 55sqrt(6)/3. I saw how you started your solution and thought 'nah, it was easy to solve and you just made it look complicated'

[@zoezulma594]

It certainly is easy to solve, if you are a math whiz. For an ordinary mortal it is rather gnarly. My approach which I find to be simpler but uses more steps uses substitutions to solve for each variable: c times a equals 300 so c equals 300 over a. Replacing c with 300 over a in b times c equals 200 you wind up with b equals 2a over 3. Substituting that for b in a times b equals 100 you get a squared equals 150. Continuing this approach to solve for the other variables you get the same answers for a squared, b squared, and c squared as you but I find transforming the sum of the squares of the three variables to the sum of the three variables to be a bit daunting. It is straight forward but 'easy' is not the adjective I would use to describe the process.

[@briant7265]

Lines 1 and 2 give, c = 2a Lines 1 and 3 give, c = 3b Substitute on line 3, 2a² =300, a = sqrt(150) = 5×sqrt(6) c = 2a = 10×sqrt(6) b = c/3 = 10/3 × sqrt(6) a+b+c = 55/3 × sqrt(6)

[@normalchannel2185]

Nah? My first approach indeed was susbtitution, but this one is also valid, and kinda easy to understand. Its the same as when we subtract eqns in Linear equation sets with 2 variables, which was taught to us in 8th grade. I, however would NOT say this question seems tough. IMO its apparent and actual difficulty is the same

[@Sci-Marvels]

I totally understand where you're coming from. The process can definitely feel 'gnarly' if you're not a math whiz. Your method using substitutions is a solid approach, and it makes sense to break it down step by step like that. Transforming the sum of the squares to the sum of the variables can indeed be daunting, even if it's straightforward in theory. Thanks for sharing your perspective-it's always interesting to see different approaches to the same problem!

[@TheBigBordedBoi]

crazy

[@wolfgangmaichen1571]

you missed the negative solution! Very easy to see. Just enter -a, -b, and -c in the original equations (instead of a, b, c) and the equations still remain valid. Thus, -a-b-c = -(a+b+c) is aso a valid solution.

[@SuperAnangs]

How come is it true?

[@normalchannel2185]

@@SuperAnangs in these cases, i like to put in values to check Basically if you have all terms in -, you can just add them and then flip the value -10 - 20 - 30 = -60. The same is -(10+20+30)

[@marionhouston2274]

@@SuperAnangs The equations to be solved have the products of two variables on the left hand side and positive constants on the right hand side. If the solutions for the variables are all negative numbers then the product of any two of these negative numbers will be positive (just as it would be if the solutions were all positive numbers). However the sum of three negative number solutions for a, b, and c would be a negative number.

[@nickcampbell3812]

I will call ab = 100 E1, bc = 200 E2, and ca = 300 E3. Multiplying E1 and E2 gives abbc = 20000 = bb(ca) = bb300 Dividing by 300 gives b^2 = 200/3 You can repeat this process for the other 2 pairs of equations. The product of E1 and E3, and subsutition of E2 gives 30000 = aabc = aa200 a^2 = 150 The product of E2 and E3, and subsutition of E1 gives 60000 = abcc = 100cc c^2 = 600. Then you just take square roots and add. Note that since all of the products are positive, the solution must be all positive or all negative.

[@DimkaTsv]

Using classic methods.... Good old substitution. Even though i hadn't mathed properly in ages already. And some people mentioned it can be even solved even simpler. It does not look like olympiad question, but rather just school homework. ab=100 bc=200 ac=300 c=200/b b=100/a c=200/100/a c=2a a*2a=300 2a^2=300 a^2=150 |a|=√150=5√6 |b|=100/(5√6)=20/√6=(20√6)/6=(10√6)/3 |c|=200/(20√6)=10√6 a+b+c = 5√6+(10√6)/3+10√6 = ((15+10+30)√6)/3 = ±(55√6)/3

[@hvnterblack]

My aproach was substitution, to solve it as 3 equasions system. You made it much easier way.

[@arunsharma-lp4vx]

I found it easiest to by by substituting and calculating each of the three variables i.e., b^2=200/3 etc - otherwise it just becomes an excercise in factorisation.

[@mikmak4228]

easiest way: ab*bc*ca=(abc)^2=100*200*300=6*10^6-------> abc=1000*route(6) then abc/ab=c=10*route(6) and so on......b=10/3 route(6) c=10 route(6)....the add them up

[@Краснаяборода-и9у]

100+200+300=600 600:2 (каждая буква встречается два раза)=300(общее число) 300-100(ab)=200(c) 300-200(bc)=100(a) 300-300(ca)=0(b) 4 класс

[@DauTheGreat]

A simple analysis shows that if the system has solutions there are two solutions: Sol. 1 with a,b,c >0 Sol. 2 with a,b,c

[@prismus6520]

* Get b and c in terms of a bc = 200, ca = 300 3bc = 2ac 3b = 2a b = 2a/3 ab = 100, bc = 200 2ab = bc 2a = c * solve for a ab = 100 a(2a/3) = 100 2a^2 = 300 a^2 = 150 sqrt(a^2) = sqrt(25*6) a = 5sqrt(6) * use value of a to get values of b and c b =2*5sqrt(6)/3 b = 10sqrt(6)/3 c = 2*5sqrt(6) c = 10sqrt(6) * add all of em together a+b+c = 5sqrt(6)+10sqrt(6)+10/3sqrt(6) =55sqrt(6)/3 ☆

[@SanePerson1]

directly from ratios of the equations: Using eqns. 1 & 2, a/c = ½ ⇒ a = c/2 ; substituting back into eqn. 3, c² = 600, c = √6 × 10 and so a = √6 × 5 Using eqns. 1 & 3, b/c = ⅓ ⇒ b = c/3 ; b = √(3/2) × 10

[@GIANFRANCOZOLA10]

It was quicker for me noticing that all the 3 multiplied are (abc)^2=600000, which means abc=1000sqrt(6). Then from the 1st you find c= 10sqrt(6), and the other two swiftly follow consequently, and then sum the 3. basically 6 steps.

[@Sci-Marvels]

Nicely done! Spotting that (abc)^2 = 600,000 and simplifying from there really speeds up the process. Once you have abc = 1000sqrt(6), the rest falls into place quickly. Thanks for sharing your efficient approach-6 steps to the solution is impressive!

[@arulo848]

a/c = 1/2 >> a = c/2 ca = (c^2)/2 = 300 >> c = sqrt(600) = 10*sqrt(6) >> a = 5*sqrt(6) bc = b * 10*sqrt(6) = 200 >> b = 20 / sqrt(6) = 10*sqrt(6)/3 = c/3 a+b+c = c + c/3 + c/2 = 11c/6 = 55*sqrt(6)/3

[@dumitru.x.5066]

From a*b=100 and c*a=300 I can assume that c=3*b. b=(√200/3), c=3(√200/3), and a=100/(√200/3).

[@Краснаяборода-и9у]

100+200+300=600 600:2 (каждая буква встречается два раза)=300(общее число) 300-100(ab)=200(c) 300-200(bc)=100(a) 300-300(ca)=0(b) 4 класс

[@RoderickEtheria]

c=10sqrt6, a=5sqrt6, b=(20sqrt6)/6 a+b+c=(110sqrt6)/6=55sqrt6/3 Signs will be the same between variables, but can be positive or negative.

[@_Tamil_gamer_]

a =12.25 b =8.17 c =24.50

[@andreyzaglodin5430]

2а=с, 2b=3a, b=3a/2, ab=a*3a/2=100, a=√(200/3)

[@susisonnenschein7034]

1. Zeile: b²=200/3 --> b=+-10/3•Wurzel(6) 1. Gleichung: a=+-5•Wurzel(6) 2. Gleichung: c=+-10•Wurzel(6) a+b+c=+-55/3•Wurzel(6) Im Kopf!?

[@Davicosban]

ca/ab = 3 c/b = 3 bc * c/b = 600 c² = 600 √c² = √600 c = 10√6 bc/ab = 2 c/a = 2 10√6/a = 2 2a = 10√6 a = 10√6/2 a = 5√6 c/b = 3 3b = c 3b = 10√6 b = 10√6/3 a + b + c? 5√6 + 10√6 + 10√6/3 15√6 + 10√6/3 45√6/3 + 10√6/3 55√6/3

[@SuperAnangs]

Oke, ab=100, bc=200 ---> b=100/a, c x 100/a=200 100c/a=200 Then ac=300 c/a=2, 2a=c => 2a^2=300, a^2=150, a=√150 a=5√6 ab=100, b=100/5√6 b=10√6/3 c=2a, c=10√6 a+b+c=5√6+10√6/3+10√6 =15√6+10√6/3 =55√6/3

[@n3m3si51]

B=200/c; b=100/a => 100/a = 200/c => c = 2a if ca=300 => a=sqrt(150). If sqrt(150) = z => c=300/z; a=z; b= 100/z => a+b+c = (400+z^2)/z = 550/sqrt(150)

[@blackcaoz]

A bit convoluted. Simply multiply all 3 equations to obtain (abc)^2=6E6 -> abc = 1000sqrt(6). Then you can find a = (abc)/bc, b = (abc)/ac and c = (abc)/ab -> (a+b+c) = (abc)[1/ab+1/ac+1/bc] = 1000sqrt(6)[1/100+1/200+1/300] =1000sqrt(6)[11/600] = 55/3 sqrt(6).

[@quigonkenny]

ab = 100 bc = 200 ca = 300 ca = 300 = 3(100) ca = 3(ab) c = 3b bc = 200 b(3b) = 200 3b² = 200 b² = 200/3 b = ±√(200/3) = ±10√2/√3 = ±10√6/3 ab = 100 a(±10√6/3) = 100 a = ±100/(10√6/3) a = ±30/√6 = ±5√6 c = 3b = 3(±10√6/3) c = ±10√6 Test: totals are all positive, so either both multiplicands are positive or both are negative. ab = 100 5√6(10√6/3) = 100 50(6)/3 = 100 300/3 = 100 ✓ bc = 200 (-10√6/3)(-10√6) = 200 100(6)/3 = 200 600/3 = 200 ✓ ca = 300 10√6(5√6) = 300 50(6) = 300 ✓ Solutions: (5√6, 10√6/3, 10√6) (-5√6, -10√6/3, -10√6) a + b + c = 5√6 + 10√6/3 + 10√6 a + b + c = 15√6/3 + 10√6/3 + 30√6/3 [ a + b + c = ±55√6/3 ]

[@kkokaki]

6:44 you forget to add the negatif solution here

[@thecrazzxz3383]

Je t'ai cramé le français

[@thunderpokemon2456]

What negative solution,there only one solution

[@igalbitan5096]

@@thunderpokemon2456 No, a² = x has 2 solutions: sqrt(x) and -sqrt(x)

[@thunderpokemon2456]

@@igalbitan5096 only one solution is suffice as a b and c are positive integer This can be prove as, b and c are positve then should also be positive. WE ARE ROOTING BOTH SIDES WE ARE NOT SQARING SO NO MINUS X

[@rainerzufall42]

Almost all comments on this video forgot the negative solution! So tragic...

[@chathuragalkandage3437]

abc(a+b+c)=ab.ca+ab. bc+bc. ca Where (abc)^2=ab.bc. ca

[@kremlinonfire]

ac = 3ab => c = 3a thus 3a^2 = 300. a = 5*sqrt(6) rest is simple substitution

[@clausanders2886]

c is actually 2a=> 2a^2=300=>a=sqrt(25*6)=>a=5*sqrt(6).. Your solution suggests a^2=100=> a=10, which is wrong!

[@kremlinonfire]

@@clausanders2886 well yea, you correct. but idea still stands I guess

[@PhysicsMathsMadeEasy3111]

Three eq multiply then find the value of abc New equation divided by given three eq Then obtained value of a,b and c

[@JAYANTI-m6f]

i did it like this here you go... ab=100, bc=200, ca=300 multiply ab to bc ab2c=20000.........................eqn 1 multiply bc to ca abc2=60000.........................eqn 2 multiply ab to ca a2bc=30000.........................eqn3 now, add eqn 1+2+3 abc[a+b+c]=110000..................eqn 4 now multiply ab to bc to ca we get , a2b2c2=6000000 takin square root on both sides abc=1000_/6 now put the value of abc in eqn 4 we get a+b+c=55_/6/3

[@duckwantbreads]

This method seems overly complicated considering that the ratios are so nice to work with, it's easy to see 2a = c immediately. From there you can work out (by subbing into the third equation) that a = sqrt(150) = 5*sqrt(6) Therefore c = 10*sqrt(6). Then subbing into the first equation you have b = 20/sqrt(6) Then it's pretty easy (by multiplying top and bottom of a, b and c by sqrt(6)) to add the three numbers up to the correct answer.

[@Sci-Marvels]

You're right! Recognizing the ratios and simplifying the problem from there makes the solution much more straightforward. By quickly identifying that 2a = c, the rest of the problem falls into place nicely. Thanks for sharing this streamlined approach-it's a great way to simplify the process!

[@rainerzufall42]

But you also missed the negative solution - 55/3 sqrt(6)

[@davidhowe6905]

For practical reasons, as each product is a multiple of 100, you can take the multiples of 10 of a,b,c so AB=1, BC=2, CA=3 (but you have to remember to convert back at the end!)

[@AyanSharma-dv1yr]

Let A be the answer a+b+c=A Dividing by abc 1/bc+1/ac+1/ab=A/abc Plug in the values 1/200+1/300+1/100=A/abc 90000/6000000 = A/abc We know by multiplying the first 3 eqs in question that a^2.b^2.c^2=6000000 abc=+-1000√6 Plug into the equation +-1000√6×3/200=A A or answer is +-15√6

[@AyanSharma-dv1yr]

Done in under 5 steps

[@Sci-Marvels]

Great approach! By dividing a + b + c = A by abc and plugging in the values, you end up with: 1/200 + 1/300 + 1/100 = A/abc 90000/6000000 = A/abc Knowing that abc = ±1000√6, you get: ±1000√6 × 3/200 = A So the answer is ±15√6. Thanks for sharing this clear and effective solution!

[@izh350by]

a=38.73, b=2.58, c=7.75 a=100/b, b=200/c, c=300/a, b=200*a/300, a=100*300/200*a, a*a=150, a=√150=38.73

[@biplobbipu7668]

b=100/a bc=200 (100×c)/a=200 c=2a. ca=300 2a×a = 300 a = +-√150

[@EvanEscher]

We can figure out that a/b = 3/2, c/b = 3/1, and c/a = 2/1. Given that, let a = x. Since we know ca=300, we know that 2x^2=300, then x=√150, which can be reduced to a=5√6. Since we know c is 2 times a, c=10√6. Since we know c is 3 times b, b = (10√6)/3 Therefore, a+b+c=(55√6)/3

[@tamhoangminh197]

To make it easilier div the first line to the third line that you have c=3b then replace c in the second line to 3b you can solve b = 10sqrt(2/3) or -10sqrt(2/3) then put the result (each case) into the first line you have result of a, and don't forget that you have c=3b above Then you can easily solve the a+b+c Do not try to make the math more complicate

[@Sci-Marvels]

Great approach! Dividing the first line by the third to get c = 3b really simplifies things. From there, solving for b and substituting back to find a and c makes the process straightforward.

[@xyki5153]

b = 1/3c 1/3c × c = 200 c^2 = 600 c = 10sqrt6 further actions are obvious

[@tomtke7351]

ab=100 eq.1 bc=200 eq.2 ca=300 eq.3 Three equations Three unknowns Using eq.1 & eq.2 eliminate b Then co-solve with eq.3 ab=100 bc=200 => (bc/2)=100 (bc/2)=ab c/2=a ca=300 c(c/2)=300 c^2=600 c=sqrt(600) ●c =10sqrt(6) ca=300 10sqrt(6)a=300 a=300/10sqrt(6) ●a =30/sqrt(6) ab= 100 30/sqrt(6)b=100 b=100(sqrt(6)/30) ●b = 10sqrt(6)/3 ●a =30/sqrt(6) ●b = 10sqrt(6)/3 ●c =10sqrt(6) verify... bc=?200 10sqrt(6)/3×10sqrt(6)=?200 100×6/3=?200 200=❤200✔️ a+b+c: ●a =30/sqrt(6) ●b = 10sqrt(6)/3 ●c =10sqrt(6) 30/sqrt(6)+10sqrt(6)/3 +10sqrt(6) A common denominator =sqrt(6)×3 =3sqrt(6) ●●a= 30/sqrt(6) = (30×3)/(3sqrt(6)) = 90 b= 10sqrt(6)/3 = (10sqrt(6)×sqrt(6)) ÷ (3sqrt(6)) ●●b=(10×6)÷(3sqrt(6)) = 60/(3sqrt(6)) c=10sqrt(6) = ((10sqrt(6))×(3sqrt(6))) ÷ (3sqrt(6)) ●●c =(30×6)÷(3sqrt(6)) =180÷(3sqrt(6)) ●●a+●●b+●●c = =(90+60+180)/(3sqrt(6)) =330/(3(sqrt(6)) =110/sqrt(6) =110sqrt(6)/6 =(55/3)sqrt(6) VERIFY: plug and chug ●a =30/sqrt(6) = 12.25 ●b = 10sqrt(6)/3 = 8.16 ●c =10sqrt(6) = 24.49 a+b+c = 44.9 =(55/3)sqrt(6) =18.33×2.45 =44.9 44.9=❤44.9✔️

[@AGNLD2009]

bc/ab = 200/100 => c/a = 2 => c= 2a ca = 300 = > 2a.a = 300 => a^2 = 150 => a = sqr(150) = 5sqr(6) c=2a = 2.5sqr(6)=10sqr(6) ab = 100 => b=100/a =100/5sqr(6) = 20/sqr(6)= [20sqr(6)]/6=10sqr(6)/3 a + b + c = 5sqr(6) + 10sqr(6) + 10sqr(6)/3 = [15sqr(6)+30sqr(6)+10sqr(6)]/3 = [55sqr(6)]/3

[@Fred-yq3fs]

yeah, super easy. A year 8 or 9 should be able to do it. ab*bc*ca=(abc)^2=100*200*300 abc=1000sqrt(6) divide by ab and you get c=10sqrt(6) divide by bc and you get a=5sqrt(6) divide by ac and you get b=10*sqrt(6)/3

[@Sci-Marvels]

Exactly, it’s a straightforward process once you break it down!

[@kpdywo848]

Excellent

[@Gnowop3]

Apply method of elimination and or substitution to get a b c is standard. Then it is easy enough to add a b c results together. Don't need to do (a+b+c)^2. Don't forget the negative answer though.

[@dongheechoi3482]

a+b+c=12/ a,b,c=2or5or5/ (a+b+c=600/abc)(abc=50,a+b+c=12)

[@azizumarov448]

b = 8sqrt(8/3); a = 1.5b; c = 2a = 3b.

[@RealQinnMalloryu4]

10^10 2^52^5 1^1^2^1 2^1 (b ➖ 2a+1). 10^20 10^5^4 2^5^5^4 2^1^1^2^2 1^1^2 1^2 ( c ➖ 2b+1).10^30 10^5^6 2^5^5^6 2^1^13^2 2^1^3^2 1^1^3^2 3^2 (c ➖ 3b+2)

[@alidennir]

10a + b = 100 10b + c = 200 10c + a = 300 +. +. + --------------------- 11a + 11b + 11c = 600 11(a + b + c) = 600 a + b + c = 600/11

[@ЕгорЗубков-у8з]

y 2 hard... c=3b; 3b^2=200; a= 3/2b; b= 10 sqrt(2/3); c=30sqrt(2/3); a=15sqrt(2/3) a+b+c=55sqrt(2/3) and do the same with negative

[@toshimakusugamo]

a = √150 = 5√6 b = √200 / 3= 10√6 / 3 c = √600 = 10√6 a + b + c = ±( 5 + 10 / 3 + 10 )√6 = ( 15 + 10 + 30)±√6 / 3 = 55√6 / 3 OR a = -5√6 b = -10√6 / 3 c = √-10√6 a + b + c = -55√6 / 3

[@mxrvvn]

Solution is ± (55sqrt(6))/3

[@oo7521]

200ab=100bc=>c=2a 300bc=200ca=>2a=3b

[@TheSandr76]

If you define from each equation the variables and solve the result, you can get the following answer: 10* sqrt(2/3) + 30* sqrt(3/2) which will be the same, but without necessity to open the third power...

[@1ucky1im]

a=10/root(2/3) b=10*root(2/3) c=30*root(2/3) a+b+c=10/root(2/3)+40*root(2/3)=root(2/3)*10/(2/3)+40*root(2/3)=55*root(2/3) a=-10/root(2/3) b=-10*root(2/3) c=-30*root(2/3) a+b+c=-10/root(2/3)-40*root(2/3)=root(2/3)*(-10)/(2/3)-40*root(2/3)=-55*root(2/3)

[@mehdimosbah3654]

a=100/b b=200/c So a=c/2 c²/2=300 c=√600=10√6 a=5√6 b=(10/3)√6 a+b+c=(55/3)√6 easy one

[@peterweusten4251]

do abxbcxac =100x200x300 (abc)^2 = 100x2x100x3x100 take square root of both sides abc= 10x10x10sqrt(6) you know ab =100 so c =10sqr(6) you know bc=200 so a= 5sqrt(6) you know ac=300 so b= 10sqrt(6)/3 adding a, b,c = 55sqrt(6)/3 done

[@TheOnlyOpie]

3 pages later... super easy, barely an inconvenience

[@ivantanaka5718]

ab/bc = 100/200 so 2a = c so a = 5sqrt(6) so c = 10sqrt6 so b = (20sqrt6)/6 sum it all and you get (110sqrt6)/6 change it to (55sqrt6)/3

[@RichardSmithers]

By the way, 110/sqrt(6) is an equivalent fraction if you don't move the radical out of the denominator. I did a couple different ways and kept getting this answer and spent a few minutes trying to figure out where I went wrong.

[@Sci-Marvels]

Good point! 110/sqrt(6) is indeed an equivalent fraction if you keep the radical in the denominator. It’s easy to get tripped up by the difference between rationalizing the denominator or not. Thanks for pointing that out-it’s important to recognize both forms!

[@yurenchu]

From the thumbnail: abcabc = (ab)*(ca)*(bc) = 100*300*200 = 6,000,000 abc = ±√(6,000,000) = ±1000√6 a = abc/(bc) = (±1000√6)/200 = ±5√6 b = abc/(ca) = (±1000√6)/300 = ±(10/3)√6 c = abc/(ab) = (±1000√6)/100 = ±10√6 ==> (a,b,c) = ±(5/3)√6 * (3 , 2 , 6) EDIT: Therefore, a+b+c = ±(55/3)√6

[@mohamadhosseins.n303]

ab^2c=20000, bc^2=60000, a^2bc=30000 sum up them give us abc(a+b+c)=110000 on the other hand (abc)^2=6000000 so abc=1000sqrt(6) a+b+c=110000/1000sqrt(6)=110/sqrt(6)=55sqrt(6)/3

[@asrielkk6969]

B = 0 A = 100 c = 200

[@ТатьянаБердеева-ж8ч]

😅 в примере умножение!

[@kamilkamil2480]

a=3x, b=2x, c=6x. Solve for x, is 10/sqrt(6). Finished

[@LordDraco2107]

a=100, b=0, c=200. a+b+c=300

[@hochmeisterulrichoffrankfu8207]

b cannot be 0 because bc is 200

[@КатяРыбакова-ш2д]

a=5V6; b=(10V6)/3; c=10V6. Достаточно заметить, что a^2b^2c^2=6000000.

[@magkostoev]

Да проще было взять извлечь корень и сложить, потом все умножить и разделить на корень из трех. ТАк даже быстрей вышло бы

[@MrGMax80]

It can be solved many much esier different ways. And negative solution is missed.

[@watch_2011]

(3) c*a= 300 => c = 300 / a (1) a*b = 100 => b = 100 / a (2) b*c = 200 => (100/a)(300/a) = 200 => 30000/a² = 200 => a²/30000 = 1/200 => a²/300 = 1/2 => a² = 300/2 => a² = 150 => a = ±sqrt(150) => b = 100/±sqrt(150) = ±100/sqrt(150) => c = 300/±sqrt(150) = ±300/sqrt(150) a + b + c = ± sqrt(150) ± 100/sqrt(150) ± 300/sqrt(150) =(1/sqrt(150))(±150 ± 100 ± 300) = ±550/sqrt(150)

[@daien1976]

I think -(55√6)/3 is also the answer.

[@rainerzufall42]

Absolutely (pun intended!)...

[@rawlenyanzi6686]

No one said they had to be integers. That was what tripped up my initial attempt.

[@joetyson3216]

2ab = bc, so 2a = c so 2aa = 300 and a^2 = 150. its plug and play from there.

[@briant7265]

Exactly.

[@AlexBaranov-u2u]

Abc=100c Abc=200a Bca=300b

[@Андрей-я7и1с]

а=с/2; а=√600/2; с=√600; в=200/√600. Сумма даст 110/√6.

[@ИльдарШакиров-и5й]

=110/√6 a, d, c = 10√6; 5√6; (10√6)/3

[@Dantido]

This seems completely wrong. You are missing the negative solutions.

[@bogdangarkusha8727]

maybe i'm mistaken looking on the surface, but minuses are ought to cancel out (if any of the variables is negative, then all are, otherwise no positive product in at least one expression), therefore positive and negative results have the same absolute value

[@duckwantbreads]

That doesn't make it completely wrong. The answer given is valid and it's obvious that if a is negative then b and c must also be negative. That means the negative answer is going to the exact same answer as given in the video multiplied by -1.

[@Dantido]

Yeah, but even then, I got (70√3)/3 doing the problem as a normal three variable system. I've checked everything and it looks okay. This way of making the problem feels fishy.

[@АлексейБигвава]

Решение правильное, но неполное. Отрицательные числа тоже подходят и ответ такой же , но с "-" минусом.

[@duckwantbreads]

​​@@DantidoI found values for a, b and c and got the exact same answer as the video. Multiple others in the comments also got the same answer. Just to be safe I also plugged it into Wolfram Alpha and it confirmed the only solution (aside from the negative version) is the one in this video. What values did you find for a, b and c?

[@captainrambell]

These kinds of person made other people scared of math and make it more traumatic

[@dd-di3mz]

I got the same answer with a=300/sqrt(600) b=sqrt(600) c=sqrt(600). I am just too lazy to simplify because calculators exist, I only find pleasure in the algebraic part, not the computational part.

[@Sci-Marvels]

I totally get that! The algebraic part can be really satisfying, and with calculators around, simplifying can feel less necessary. It's great that you enjoy the problem-solving process-sometimes that's the best part!

[@Madgearz]

ab =100 bc =200 ca =300 b =100/a b =200/c a/100 =c/200 a =½c a =100/b a =300/c b/100 =c/300 b =⅓c a +b +c =½c +⅓c +c =c(½ +⅓ +1) =¹⅙c 100 =ab =(½c)(⅓c) =⅙c² c² =600 c =±600⁰•⁵ =±10(6⁰•⁵) a +b +c =±¹⅙10(6⁰•⁵) =±55(6⁰•⁵)/3

[@yavuzsensei6458]

ab=10a+b=100 bc=10b+c=200 ca=10c+a=300 +_______________ 11*(a+b+c)=600 a+b+c=600/11 very easy indeed 😂

[@tacemus]

Interesting. Thank you very much.

[@КирилСаріогло]

a=2, b=50, c=150, a+b+c= 202.

[@peak2313]

เอาโจทย์คูณกันแล้วหารากที่ 2แก้สมการจะง่ายกว่านี้เยอะเลยครับ

[@TungA-r5q]

(abc)^2=6x10^6 abc=+-/6x1000 So easy get a,b,c

[@Linkedblade]

Didnt have to expand the trinomial squared. Could have just taken the square roots on your first three lines and summed them