Spain Math Olympiad Question | You should be able to solve this!

[@sudhirjoshi7782]

Just multiply ab and bc and divide by ca. You get b². Similarly, a² and c².

[@ClickBeetleTV]

This is what I did

[@marshan1502]

Nice

[@dileepmv7438]

she was travelling from India to srilanka via Dubai Germany Mexico Japan Singapore

[@SamHalls2015]

The problem is that 2,000,000, which is the product of ab and bc, is not divisible by 3, like ca is.

[@ClickBeetleTV]

​@@SamHalls2015That is a feature, not a bug

[@SotirakisPeklivanas]

There is a simpler method. Just find the square of a, b and c then add them together, by taking two equations and substituting into the third to find the square. However I do like your method.

[@memeplex9096]

You right: a^2+b^2+c^2=ab*ca/bc+bc*ab/ca+bc*ca/ab=1000*3000/2000+2000*1000/3000+2000*3000/1000=24500/3

[@turtletalks1]

ab=1000 so b= 1000/a bc=2000 so b= 2000/c 1000/a = 2000/c 2a=c ca= 3000 So 2a^2 = 3000 a^2 = 1500 Now factor this into original solutions to find b^2 and c^2

[@olddog57]

Didn't see your comment before, otherwise wouldn't try: I even did virtually the same in my mind and commented too

[@vranix]

Followed the same way of solving it. No square roots needed 😛

[@adriensobieski979]

ab x bc / ca = b2 = 2000 / 3 bc x ca / ab = c2 = 6000 ca x ab / bc = a2 = 1500 And then just sum.

[@rufescens]

That's also the approach I found.

[@MrMichelX3]

I did it without using a constant "k", but by multiplying the 3 equations each others. You can multiply by example ab x ca = a²bc=2000a² ...etc.

[@ulisses_nicolau_barros]

Very nice question. I've liked a lot of those problems from Olympiads. God keep blessing you. One of the best Math channel in KZbin! 😄

[@leotrieu84]

5 seconds solution: ab x ca = a^2 x bc = a^2 x 2000 = 1000 x 3000. So a^2 = 1000 x 3000 / 2000 = 1500 Do the same with b^2, c^2

[@pcrtwentekanaal8458]

Indeed that is how I did it too. More elegant and less ink needed!

[@prochetoshsanyal]

This is how I did it. So much simpler

[@hsjwvs]

Why did a^2 become 1000, im confused TT

[@luca48-ex3pj]

I agree with people saying "a lot of work for such an easy, immediate solution" obtained with (cyclic) product of two expressions divided by the remaining third ...

[@carlowood9834]

Imagine you have an aquarium where the surface of the side is 1000, the bottom surface is 2000 and the front is 3000. What is the square of the longest distance a fish can swim in a straight line? The answer is going to be 1000 times larger than what we get using 1, 2 and 3. So, 1 m^2 side (depth x height) and 2 m^2 bottom (width x depth): it's twice as wide as high! And the front surface is 3 m^3 (width x height) is thus 2 height^2 --> height^2 = 3/2 m^2, and width^2 is four times that: 6 m^2. Since the side is 1 m^2, we have depth^2 x height^2 = 1 and thus depth^2 = 2/3. Add it all up: 2/3 + 3/2 + 6 = (4 + 9 + 36) / 6 = 49 / 6. And the final answer is 1000 times that is 24500 / 3

[@olddog57]

Too complicated solution for a task solvable without a pen, just in one's ming (I just did): 1. Divide 2nd equation by 1st getting c=a/2 (or a=2c) 2. Substitute in the 3rd equation first c by a/2 getting at once a²=600 3. Substitute in the 3rd equation now a by 2c getting c²=1500 4. Divide 3rd equation by 2nd getting a/b=3/2 or a=3/2b and substitute with this a in the 1st equation, getting b²=2000/3 5. Altogether a²+b²+c²=1500+6000+2000/3=24500/3

[@mr13lack]

yeah same my method

[@pedrocalleja123]

@@mr13lackc=2a

[@nicolasbrauner4898]

I think you have a typo at a^2 here. should be 6000 to make sense

[@rodnymachado3252]

a²=1500

[@legendarysquad8778]

Bro in this way i find 13500 ans but how?

[@prabhudasmandal6429]

The other way is, take 1000 as common and set it aside for final stage. So, ab=1, bc=2, ca=3.squaring, these become, a^2b^2=1say(A), b^2c^2=4,say, (B) c^2a^2=9.say,(C) now, multiplying the given 3 datas we get ab×bc×ca=1×2×3 or a^2b^2c^2=6, say,( D) . Now, dividing( D) by (A), (B) and(C) we get, c^2=6, a^2=3/2and b^2=2/3.so, a^2+b^2+c^2=49/6now, import the common 1000 to multiply it and get 49/6×1000=49/3×500=24500/3.Ans.

[@brandonk9299]

It is much easier to do the division. bc/ab= c/a=2000/1000=2. Now multiply that by the last so c/a x ac = 2 x 3000= c^2 = 6000. We know that c=sqrt(6000). So a=3000/c= 3000/sqrt(6000) and b = bc/c=2000/sqrt(6000). These can now be squared and the radical disappears: a^2+b^2+c^2= (3000)^2/6000 + (2000)^2/6000+(sqrt(6000))^2= 9000/6+4000/6+6000= (13000+36000)/6 49000/6= 24500/3 = 8166 2/3

[@cigmorfil4101]

When given some statements regarding somethings, but the answer requires somethings squared then the actual values to the somethings are likely not be required to get the solution, just what those values squared are. Which in this case leads to the question of how can I get a^2, etc from the equations given. A quick look gives that[1] ab × ca / bc = a(bc)a / bc = a^2 Now just manipulate the numbers corresponding to the pairs of multipliers and the value of a^2 pops out without needing to work out a. Similarly for b & c. [1] I suspect this is the method that the writers expected to be used for the quickest, easiest solution.

[@month32]

I think the reason for the constant κ is there as an example tool, in case you need to use this method for other, harder questions.

[@sriramrc2681]

Nice thought bro…

[@m.caeben2578]

Well, I took a long detour. Here it is my reasoning: 1. Let x≡1000, simply to have, ab = x, bc = 2x, and ca =3x. 2. Notice (a+b+c)^2 = a^2+b^2+c^2 +2(ab+bc+cd) = a^2+b^2+c^2+12x. 3. To solve for a+b+c, notice that by multiplying the original system by the remaining variable, one has abc=xc, abc=2xa, abc=3xb, and to get a+b+c, we multiply the equations so that the rhs's have the same factors, to the a,b, and c. Aka: 6abc = 6xc, 3abc = 6xa, 2abc = 6xb, which then by summing, 11abc = 6x(a+b+c) 4. To solve for abc, we multiply the original equations by each other => (abc)^2 = 6x^3. 5. Substituting everything as we got them: a^2+b^2+c^2+12x = (a+b+c)^2 = (11abc/6x)^2 = 121 x/6 => a^2+b^2+c^2 = (121/6 - 12)x, with x = 1000.

[@dienodecol1808]

Hello nice exercise.Idid like this ; a/c=1/2 , b/a=2/3 , b/c=1/3 , b/a*b/c=2/9 b²/3000=2/9 then b²=6000/9 etc......

[@sergiykanilo9848]

a^2 =ab*ca/bc b^2 =ab*bc/ca c^2 =ca*bc/ab

[@ryun-.-4164]

Gọi lần lược 3 phương trình từ trên xuống dưới là (1) (2) (3) (1) ÷ (2) được a/c = 1/2 => c = 2a. Thay vào phương trình (3) được 2a^2=3000 a^2=1500 . c=2a => c^2=4a^2=6000 Lấy (2) ÷ (3) ta được b/a = 2/3 => b = 2a/3 => b^2 = 4a^2/9 = 2000/3 Từ đó được đáp án là cộng 3 số lại

[@soloking2469]

We in india used to slove this type of questions in class 10th..

[@ChitrakGupta]

A^2 = (ab x ca)/bc.. calculate b^2 and c^2 similarly…

[@memoire4953]

Just multiply the three equations and get a^2 b^2 c^2 = 6 000 000 000 Since a^2 b^2 = 1 000 000 b^2 c^2 = 4 000 000 c^2 a^2 = 9 000 000 a^2 = 1500 b^2 = 2000/3 c^2 = 6000

[@stormlivin]

just divide a in ab i.e b = 1000/a, and in bc i.e b = 2000/c… now multiply those two equations you will get b , and then solve it for c and a by using the value of b.

[@schlingel0017]

I just solved it like a normal equation system and found values of a b and c

[@spencer7404]

This is literally a 4 grade test in China. I saw it in my son's homework once.😅

[@sergefrechette8995]

Isnt it obvious that this is two integers 7 x 5 as the logical outcome unless x and y are specified other . The second statement supposes the obvious, that the first statement x and y are integers.

[@Miss_miss-u3b]

Why its so complicated???? In 1:31 you already got solution!!! But you make it into rocket science 😕

[@bhchoi7057]

Is it almost trivial by using log and add 3 equations?

[@sansanichalexandroff8913]

I am still trying to figure out what kind of accent is that? Where is she from?

[@aek5797]

Why not solve ab=1000 for a; bc=2000 for c; put it in ca=3000 and from there it’s a cake walk to a, c and a^2+b^2+c^2=? ?

[@frankzhou8617]

This should be a primary school question.

[@kennethkan3252]

ab=1000 bc=2000 ca=3000 a:b:c=3:2:6 abxbc=1000×2000 ac(b^2)=1000×2000 ac=3000 b^2=(1000×2000)÷3000 b^2=2000/3

[@anvu6631]

bc/ab=2000/1000. so that c/a= 2. and then solve 2a^2=3000 easily

[@vecuivious]

What pen are you using??

[@Xebtria]

Well I did the complicated route. STEP1 a=1000/b c=2000/b c*a = 3000 therefore (1000/b)*(2000/b) = 3000 2000000/b² = 3000 2000000 = 3000b² b² = 2000000/3000 STEP2 b=1000/a c=3000/a b*c = 2000 therefore (1000/a)*(3000/a) = 2000 3000000/a² = 2000 3000000 = 2000a² a² = 3000000/2000 STEP3 a=3000/c b=2000/c a*b = 1000 therefore (3000/c)*(2000/c) = 1000 6000000/c² = 1000 6000000 = 1000c² c² = 6000000/1000 STEP4 x = a²+b²+c² x = 3000000/2000 + 2000000/3000 + 6000000/1000 x = 3000/2 + 2000/3 + 6000/1 x = 9000/6 + 4000/6 + 36000/6 x = 49000/6 x ~ 8166.67 well I missed dividing by 2, but it is the same result in the end.

[@prudvi7797]

Actually by direct substitution also we can get its a=√1500. and b=2/3(√1500). and c=2√1500 All values satisfy every equation and i got the answer 73500

[@ShriABCD]

But guessing that a equals root 1500 is not that easy

[@prudvi7797]

@@ShriABCD i calculated and said this not saying without doing

[@ShriABCD]

@@prudvi7797 Ok 👍

[@prudvi7797]

@@ShriABCD did u try it the values alo satisfied ab bc and ca too

[@thurmanmerman2720]

In my country this test is done by infants who have yet to talk. Then the teacher chops them in two with a rusty knife and dances on their grave.

[@manjurhasanchisti2264]

A complicated solution.

[@legendarysquad8778]

Maybe the answer is wrong cause by dividing with each other, i find 13500 ?!

[@unnamed4218]

Is not working with the formula of (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca ? just asking :D

[@rohitpandey8681]

i was thinking the same

[@t-o9844]

это можно прекрасно решить без введения дополнительной переменной

[@VivekPandey-sk9ln]

I don't think so that this type of easier question is from Olympiad of Spain

[@Денис-любительхимии]

bc/ab=c/a=2, CA=3000

[@FinnWalTD]

Very long solution of simplest linear sistem.

[@BP-gn2cl]

Very lenghty solution

[@ciaucia156]

=(1*3/2 +1*2/3+2*3/1)*1000

[@someone._.5333]

Interesting problem

[@hottuamanullang8046]

Is that somebody get the answer for 8500?

[@dannuttle9005]

This was my second math puzzle of the day and I got them both right. I am a complete idiot, so this is a red-letter day! Huzzah! Bring me my hat and walking stick, I shall proceed to the club to boast!

[@RobScha]

Very very complicated way...

[@zsboya]

ab * ac / bc + ab * bc / ac + ac * bc / ab = a^2 + b^2 + c^2 --> /1000 --> 1*3/2 + 1*2/3 + 3*2/1 = 9/6 + 4/6 + 36/6 = 49/6 --> a^2 + b^2 + c^2 = 49/6 * 1000 = 49*50*10/3 = 24'500/3 .

[@in_ui5912]

abbcca(1/abab+1/bcbc+1/caca)

[@Aines_Rota]

I solved this math problem with ratios

[@RUDYFERNANDOGONZALEZESCOBAR]

Super like desde Guatemala

[@robertb3214]

She made this far too difficult.

[@kingsukbandyopadhyay9662]

This is very simple mathematics.

[@evrankorkut8915]

tenkyou

[@ВячеславКозлов-щ9д]

Решение через жопу ) Вы не ищете лёгких путей, в комментариях много раз уже написали более простой и короткий способ решения.

[@frosinabrahja1792]

500

[@rajug5709]

24500/3

[@victorgorelik7383]

small variation: kzbin.info/www/bejne/Zl6ylYKlaLmojLc

[@ناصرالعتيبي-ك1و3ب]

a=40 b=25 C=80 مجموع ناتج المعادله 8625

[@ناصرالعتيبي-ك1و3ب]

الناتج 8625

[@MikhailGromov33]

ac=3200?

[@ناصرالعتيبي-ك1و3ب]

40^80=3200

[@Th0Rx666]

You could just guess i guess Xd

[@progameronly9852]

Anwser is 1000000+0+9000000= 10000000 easy

[@JPTaquari]

sorry, my method is easier, Hugs From Brazil!!!!!!

[@alster724]

Easy

[@jmkeupp]

Huh?

[@lv70rouge]

literally 3rd grade math can solve this faster and easier

[@WeCareTV-z1r]

Your answer is wrong it will be 12500/3

[@babyghost4434]

great solution

[@srinivasanramachandran6528]

simpler is (ab)*(bc)/ca=1000*2000/3000 = 2000/3 or b^2=2000/3 similarly a^2=1500 and c^2=6000 and answer comes by adding

[@babyghost4434]

@@srinivasanramachandran6528 also great solution sir

[@LicensedTeacherFor4y-7y]

💛💖 Thanks for being important material during my whole lives 💛💖 🟥 🟧 🟨 🟩 🟦 ⬜️ ⬛️