How To Solve An MIT Admissions Question From 1869
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- Опубліковано 4 жов 2024
- The Massachusetts Institute of Technology (MIT) is one of the top ranked universities in the world. This question appeared on its admissions exam nearly 150 years ago. "The perpendicular dropped from the vertex of the right angle upon the hypotenuse divides it into two segments of 9 and 16 feet respectively. Find the lengths of the perpendicular, and the two legs of the triangle." The video presents a solution.
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We're all going to MIT guys, great work.
Me too bro!! Doctorate for something in no time!
Ok, maybe not bro, sorry.
Richard Wendt no shhhhh it's happening
I had to look at the sketch because I didn't really know what a perpendicular was (not an english native)...am I still admissible? :X
I thing all languages is the same word
MIT would be very crowded if this still existed
It was only one question. I'm sure the test was harder in general, and timed. My method was longer.
It was from 150 years ago. Now MIT is just integration bees and calculus.... and calculus.
OneWeirdDude, how longer?
@@markiyanhapyak349 Using the Pythagorean Theorem.
Mh.....hm; You have to use it, but later..... . How did You get to the theorem?
Well cool looks like I can just drop out of middle school and go join MIT now
you mean 150 years ago?
Do y'all understand the concept of a joke?
I would like but there's 420 likes at the moment. this is my like lol
same lol
ua-cam.com/video/Ns28oYxGURA-/v-deo.html I’d love to know your thoughts on this
I solved it differently, using the pythagorean theorem for the whole solution. I labelled the unknown legs as x,y and z. The large triangle formed by combining the 2 smaller triangles is represented with this equation: x²+y² = 25². The triangle to the left is represented by : 9²+z²=x². The 3rd triangle is represented as: z²+16²=y². Now we have 3 unknowns and 3 equations so we can solve by substituting the values of x and y into the first equation and that looks like this: z²+81+z²+256=625 and this reduces to 2z²=288 and then z²=144 so z=12. From here it is easy to solve for x and y.
same
same here Mike
I also used this method, because I didn't know or had forgotten about that similar triangle rule that he used in the video
same here!
yes, I used this method too
I used Phytagoras for every step because I'm a fan:
9²+h²=x²
16²+h²=y²
x²+y²=(9+16)²=25²
9²+16²+2h²=25²
2h² = 288
h² = 144
h = 12
16²+12²=y²
400 = y²
y = 20
25²-20² = x²
x² = 225
x = 15
+
Not as simple but just as effective.
This was how I tackled it too. It's nice not having to remember that the triangles are similar when doing it this way.
I was going to post how I got the correct solution, and then I read your comment and said, "Hey, that's how I did it!"
From where did u take 25??
Once you realize it’s a 3-4-5 triangle, you can just sum the segments to get 25 for the outer triangle, see that it is a quintuple and get 15 and 20.
Solved it in the exact same way
There's not enough information at the start to realize it's a 3-4-5 triangle, you only have the length of the hypotenuse. Not all triangles with a hypotenuse of 5 are 3-4-5 triangles. It just so happens that this one is.
Engineers: *furious quarrels about using addition vs ratio for solving this*
Craftsmen: brb gonna get my ruler
@@wvucyclist1 I believe the perpendicular bisector identified at 1:12 tells us that all 3 triangles are similar, therefore as soon as we observe that the inner two are 3-4-5, we know the outer one is as well.
@FreymanArt exactly. One side could be a surd, or it could be 7-24-25
people are like "Wth is this MIT?" "It would be so crowded if it were this easy.."
yo it was 200 years back then, what do you expect?
Victor Nguyen You realize the Pythagorean theorem and Euclidean geometry have been around for much longer than 200 years, right? An ancient Greek could have passed this test.
@@micahgoldson1253 but was everyone learning it back then?
Tomás Beltrán Supposedly. American schools started teaching geometry in 1844. 25 years before this test question.
@@micahgoldson1253 interesting, thanks
Tomás Beltrán No problem!
I was able to solve it. Now I can clear MIT admissions.
Just need to go 150 yrs back in time.
Has anybody got a time machine?
I've have a truly marvelous idea for a time machine for which this comment is too limiting to contain.
@@seitch1 😮
After all, you're a prospective MIT applicant. Build your own for god's sake.
@@sudoLife Exactly and then go back in time to present your invention.
Wait if you make a time machine that means you can take admission at the present time.
My small brain hurts with all this
😂😂
Should've been born in 1851...
Bill Kong lol
Bill Kong True that.
Bill Kong but no internet for u... sad face
If you can't solve this simple math problem, you have no chance at understanding the rigorous statistical methods needed to document disease progression and whether YOUR "cure" really works on a large group of either patients or lab rats. Yup, snowflake, it's a tough world!
Wow... so complicated..... 0_0
This is certainly one of my favorite puzzles in this channel, i discovered many things here. Apparently, in a right triangle, when you draw the perpendicular from the right angle to the hypotenuse, it divides the triangle into 2 similar triangles, both of which are also similar to the main triangle.
Yeah that is a theorm
If u draw a perpendicular from the vertex of a right angle of right triangle to the opp side, then the two triangles formed on sides are similar to each other and to the whole triangle.
That fact is at the core of Euclid's proof of the Pythagorean Theorem.
Why wouldn't p = 2sqrt(39)i?
In the entire large triangle, call the left leg x and bottom leg y
9^2 + p^2 = x^2
p^2 + 16^2 = y^2
x^2 + y^2 = (9 + 16)
Plug in for x and y
(9^2 + p^2) + (p^2 + 16^2) = 25
p = ±2sqrt(39)i
@@maxhagenauer24 9+16 should be squared
Finally, the first ever question from Mind your decisions, that I did on my own. Thankyou for your service, my rusted brain has started working again.
You can easily find a sequence of equations -> 9^2 + p^2 = a^2 , a^2 + b^2 = 25^2 , (b^2=25^2-a2) , 16^2 + p2 = b2. 16^2+p2=25^2-(9^2+p2) => p=12, using the pythagorean theorem, you can find the legs - a,b.
my first thought was same
I just did this and i am not even born yet.
Speed Demo I did this one back in 1869, what u wanna say brah.
I don't want to imagine how you got a device to post the comment
I just did it and My father hasn't even been born yet
so, clean it up
Speed Demo they teach us this in 10th grade in 🇮🇳
simply label the sides as 'a', 'b' and the perpendicular as 'c'. Then solve the system of questions:
1. a^2 = 9^2 + c^2
2. b^2 = 16^2 + c^2
3. a^2 + b^2 = (9+16)^2
use substitution (1) and (2) in (3) and get c = 12, and then a = 15 and b = 20.
none of this scale up and down nonsense. simply use the Pythagorus theorem from grade 5.
Yep, pretty much what I did but without a calculator, there's some bigger multiplication that takes a few seconds more to do. (for me anyway)
not really.. its just how easily you can recall a method for doing it
Recognizing the similar triangles and seeing the 3-4-5 triangle pattern is the fastest way to solve the question. Yes, the Pythagorean theorem works because that's how you prove this works, but on an exam, it takes too long, and someone with solid geometry takes seconds to see the patterns and come up with the solution.
It makes perfect sense to well... smart people.
aliesneo can you message me how you used that particular method.... bit lost. Thanks!
The length of the perpendicular segment (altitude) is the 'geometric mean' of the two defined line segments -- a useful tidbit from geometry.
It arises from similarity, which can be used to prove the pythagorean theorem itself.
For the last step - the large triangle is also similar to the 2 small ones, thus the 3-4-5- triangle, and since the hypotenuse is 25 the legs are 15 and 20
I did it a different way, I called the legs a and b, and applied pythag. 3 times to get
a^2 + b^2 = 25^2,
p^2 + 9^2 = a^2,
p^2 + 16^2 = b^2
Adding equations 2 and 3 gives 2p^2 + 9^2 + 16^2 = a^2 + b^2 = 25^2, rearranging gives p^2 = 144 and p=12, then a and b from there.
yeah me too mate. But this method is longer and requires more calculation than the one presented in the video.
I also did it this way and I believe it would be the most intuitive approach to those that may not know or remember all the geometry theorems offhand. Pythagoras shouts out at you in this one.
Welcome to MIT guys. ;-)
yes, I did it this way too hehe. I guess it is "longer" but it works.
i just used set of three equations based on Pythagorean theorem...
same here
Same here too... Took like 3 seconds... Lol
you can only do that once you get the original "p" length ... if you can figure that out as he did then yes, your method is much simpler but the question is, how did you get "p"?
David Beecroft easy: 81+p²=a², 256+p²=b², a²+b²=625
If we add first two equation and then we get:
337+2p²=a²+b²=625
2p²=288
p²=144
p=12
Same here. Took less than 5 seconds
Too easy for high school students.
And that's why this is from 150 years ago.
does that mean our education system has improved in those 150 yrs?
geometry has been a very advanced branch of mathematics since centuries, even millennia.
I never would have thought of both triangles being similar.
One doesn't need to. If you know your 1 to 10 multiplication table, and the pythagorean theorem, you can solve this.
This was super easy for me it was one of the basic questions in a math book in *INDIA*
And that too of 6th grade
Yeah true
It was in indeed an easiest question
Isn't it similar to proving Pythagorean theorem?
@@togbot3984 yes it is
@@pablosanchez6877 If it's old it doesn't mean that it should be everywhere and please don't take me as someone who only thinks about stereotypes. I don't mean to offend you
So you want to say you'll gonna top that MIT admission test for 6th graders?
Three unknown variables, three equations relying solely on the Pythagorean Theorem.
a^2+b^2=(9+16)^2
16^2+p^2=b^2
9^2+p^2=a^2
+bakeslash: It is really much simpler than that Since all the triangles must be similar, the solution for p is simply 16/p = p/9. p = 12 Which is what your long process gives you.
I solved the problem with a 3-equation system, I think it's clearer that way.
Javier Sds same
this is actually significantly easier than any question I was asked at interview or did on the required exams
Geometry tends to come easier to those who think visually compared to those who think in words.
It is reassuring to know that the mathematical skills of those going to academia have actually progressed a lot for the last 150 years.
a2+b2=25x25
9x9+p2=a2
16x16+p2=b2
Substituting a and b in terms of p
81+256+2p2=25x25
P2=144. P=12
Substituting value of p
9x9+p2=a2. a=15
16x16+p2=b2.b=20.
9^2 + 16^2 + 2p^2 = (9+16)^2 = 9^2 + 16^2 + 2*9*16 = 9^2 + 16^2 + 2*(3*4)^2, so p = 3*4 = 12 (saves a bit of the calculation effort)
This was my method also
I solved this using the fact that the radius of outer circle is in the middle of the hypotenuse, so it's 12.5 from the middle to every vertex of triangle. so you get another triangle with sides 12.5, 16-12.5=3.5 and p, and you get the p using pythagorrean theorem. once you get the p, similarly you get remaining two legs :)
That was creative:)))
Nice!
thanks everybody :)
Yeah, I did the same, but then with the video explanations I've just realized that this was way too more complicated.
another really nice solution!
Can I go to MIT now?
I did not use similar triangles at all. I only used Pythagorean Theorem. using the equation x^2+9^2=A^2
x^2+16^2=B^2 and A^2+B^2=25^2. 2x^2+16^2+9^2 = 25^2(after some algebra and a substitution).
that was what I used but I got the perpendicular to the hypotenuse = 0.
I did the same and got the correct answers!
Similar triangles is waaaaaay easier, it got me 30 sec to solve it. How much time did you have to use to do it with that?! I mean "after some algebra and a substitution". Pluss, there is always the 3-4-5 triangle and its multiples!
can ulker yup. you are totally correct. similar triangles is way easier! should have done it that way! lol
can ulker öklid kurallariyla direk cikiyor asil 10 sn bile sürmüyo benzerlik yine daha uzun suruyo.
I just used the Pythagorean Theorem with three equations. The easiest way to do this in my opinion is:
p^2 + 256 = a^2
p^2 + 81 = b^2
a^2 + b^2 = 625
By substitution: 2p^2 + 337 = 625, so p=12
If p=12, a=20 and b=15
Thus actually is the first sum of this channel that I’ve solved on my own. Finally I’ve improved. Thanks for these sums. I tried both Pythagoras theorem as well as similar triangles. Both work
not big deal, childeren can solve this before they go to high school
I would say quite a few high school graduates would have trouble with this.
... isn't it just -30
mz atmaca Sadly, your kids are the kids of a troll who thinks he knows everything. I predict they will be in high school for 5 years.
Narata haahaa... I know i have exaggerated... But still this question is not exclusive to MIT, not special. We used to solve that kind of prob at math exams in high school
mz atmaca ehh your almost right im a freshmen learned this last semester
Huh... I did it by Pythagorean theorem (system of 3 equations). But similar triangles work too... evidently.
Going out on a limb here, seems like MIT has updated their acceptance criteria over the past ... 150 years.
I did it that same way (Pythagorean theorem and system of 3 equations), but I messed up somewhere and I got something different. I must have messed up the math somewhere. Curse addition and subtraction!
I used this method also.
So did I.
so did i
ikr a lot easier
I did it easily using only Pythagorean theorem. I just didn't remember what it happens when you drop a perpendicular from the right angle in a right triangle. But now that I saw it I can even prove it with two equations using the angles.
Pythagoras. Protagoras was an philosopher :D
LOL you're right, that was my cell phone giving its opinion.
Lithus17 You looked at it and knew the perpendicular was 12? amazing
I didn't remember that tidbit either, but if you call one of the angles formed by dividing the original 90 degree angle theta, then the third angle of the new, smaller triangle must be 180 - 90 - theta = 90 - theta. Since the original right angle was cut by the perpendicular, the remainder of that must also be 90 - theta. And anyway you solve it, the last original angle of the large triangle must be theta. Now you have 3 triangles, all with angles of 90, theta, and 90-theta, and they are similar. Not remembering the rule led to a bit more work, but the same ultimate solution as our host.
I used “a” for the dropped perpendicular, “b” for the hypotenuse for the right triangle with side 9 and “c” for remaining leg. There are 3 right triangles with 3 unknowns which can be solved by elimination. a=12, b=15, c=20.
16^2 + z^2 = x^2
9^2 + z^2 = y^2
y^2 + x^2 = 25^2
16^2 + 9^2 + 2z^2 = 25^2
z = 12
And so in, didn't bother with the simpler actions
Puzzlesolverino?
Yep - simpler. No "special set of triangles" rule to remember
I did it algebraically as well--all that was needed was the Pythagorean Theorem.
i have do the same
This is the best answer. If you can solve it with Pythagoras, don't bother with anything else.
I went way a somewhat different way and solved the two sides first before the perpendicular
Labelling the 2 sides as a and b, and the perpendicular p, you have:
p^2=a^2-81=b^2-256 leading to:
a^2=b^2-175, substituting into a^2+b^2=625 you have:
2b^2-175=625, so b^2 is 400, putting back into a^2=b^2+175, a^2=225, leading to a=15, b=20
Going back to p^2=a^2-81, we have p^2=144 so p=12
I like this - I used it - solving simultaneously has wider applications.
I did it by:
a^2=81+c^2
b^2=256+c^2
a^2+b^2=625
So, 81+c^2+256+c^2=625
=> 2c^2=288
=> c=12
Putting this back in the top equations gives a=15 and b=20
Same as yours really.
You don't have to know that two internal triangles are congruent.
+manudude: My, that got kind of tediously long didn't it?
First, notice, and prove very simply that the 3 triangles are similar. Thus all the sides on all the triangles are proportional in length.
Thus it obviously follows that p/16 = 9/p. Thus p = 12
That is the entire solution, it is really just that simple
As Einstein said, the main hallmark of genius is simplicity.......
So less is more impressive than more........
Well i just looked at 9+16=25 => the too other sides have to be 15 and 20 beciuse it is simply the 3,4,5 triangle *5.
Almost, it could be a 7,24,25 triangle so you need to show it works for one or the other. Only a little more effort though. I too assumed the 9+16=25 was a clue.
my first guess was in fact 7, 24, 25 b/c i failed to see the multiple of 3,4,5! But then you end up with one triangle with hypotenuse 7, and a leg of 9, lol, so I scrapped that and kept looking until i found the 20 and 15.
It works that way easily and intuitively, because it is a perfect square for both of the legs. If you noticed, you were still doing similarity. So if you do not have a perfect square for both legs, you'd end up with a different solution.
multiple of 3,4,5 or 3,4,120?
ModProg wow, i haven't thought of that, you must be really smart recognising the 3,4,5 triangle and using it correctly
It can also be solved by simple equations:
Y^2-x^2=175
X^2+y^2=625
That's it, just solve these two equation to get the answer if you don't have any idea about similarity of triangles.
how
@@jambasjoe6045 the left leg will become the hipotenuse to the 9-p - hip1 triangle and the right leg, the hipotenuse to the 16-p- hip2 triangle. Than, you take the results for the previous pythagorean theoram equations and use them in the triangle hip1-hip2- 25 triangle, and teh only incognite will be 'p'
3 equations. 2 for the smaller triangles and one for the overall. Substitute and solve.
@@sain8827 Right! That is how I solved it.
No similar triangle stuff necessary.
Moreover, this method is general.
Dude u can find 12 from Euclid's formula;
P² = 16.9 With this you can find that more easily.
İrem Lindemann which has been derived from similarity
yes i did that
Thats what I did
i agree with you Guardian
Trumpet-Lover exactly! people are spitting out these formulas and equations when you don’t even need to use the pythagorean formula much less Euclid’s formula. Just the 3,4,5 triple
I quickly realized that 25 = 5*5, so the triangle and to be a scaled version of the 3-4-5 triangle and so are all the triangles in this problem, thus the length of the perpendicular works out to be 12, with legs 15 and 20.
Maybe I just got lucky ;-)
That's the first guess I made. Followed through on my assumption and it panned out.
And that is why the more challenging problems are slight variations to this one where that actually gets you the wrong answer.
just a guess but I doubt that this would pass examination - there are infinite right angle triangles with an hypotenuse of 25...
but not an infinite amount that cuts the hypotenuse into 9 and 16, so for this case that was perfectly valid. If it was a different but one of those other similar problems, a simple check of the solution and logic test would reveal if something was amiss.
Cool! Folks in the construction trade might've seen that right away, too. They know the 3-4-5 rule and simply expand it as necessary to get right angles for houses and such. I *should've* seen that but I went 3 equations, 3 unknowns.
So much hard thinking for a problem which can be solved with raw force as set of three Pythagorean equations with some trivial substitutions. I've just did it in a notepad. (no calculator needed, square roots for 144, 225 and 400 are very obvious... sure, 16^2 and 25^2 are a little annoying, but it can be done from top of the head as well)
If anyone wonders (you shouldn't), here it is:
p^2 + 9^2 = a^2
p^2 + 16^2 = b^2
a^2 + b^2 = 25^2
(p^2 + 9^2) + (p^2 + 16^2) = 25^2
p^2 + 81 + p^2 + 256 = 625
2p^2 + 337 = 625
2p^2 = 288
p^2 = 144
p = 12
144 + 81 = a^2
225 = a^2
a = 15
144 + 256 = b^2
400 = b^2
b = 20
1) it's not hard thinking, it's taught in 10th grade geometry
2) a big part of math is finding easier ways to do things rather than just relying on brute force
Simultaneous equations.
Not hard thinking, a little bit of consideration finds a so much more elegant and simple approach. Don't feel bad. I rushed to a solution using pythagorean identities and simultaneous equations and missed it too. Not a wrong answer, but the similar triangle comparison is so much better!
Its not a test, this is for fun snd education. Explore!
He skipped a step, before he can use similar Triangles he needs to show that the angles are equivalent to each other like name top left Alpha and bottom right Beta
We know Triangles are 180 degrees, and that is a right angle so you have 90, alpha and 180 - 90 - alpha = 90 - alpha.
like wise for the bigger triangle, 90 + beta = 180 so the other angle is 90 - beta and
we know 90 - alpha + 90 - beta = 90 which is 90 = alpha + beta and sub the equivalent equation into other angle alpha or beta to show they are similar then you can use the similar triangle.
sub one angle into another and you have equivalent triangles, without this step the answer is wrong. Especially in an exam, you can't make assumptions. Cause the marker would be saying how did you get to here from here..
It's like the Discrete Maths exam, Maths markers are extremely picky, show that a path is a cycle,simple cycle or nothing a) ABCDAB. You have to prove to them it works that's how come Computer Science and Maths goes hand to hand, it's ok to show a Program does work but you have to PROVE it works.
If you say it's a cycle, you just received one mark cause you ONLY answered one part of the question, the question ask to demonstrate all 3. You missed 2 easy marks cause you made assumptions.
Latex would be handy since this looks dirty.
Becareful bud, his answer isn't complete, he wouldn't receive full marks on the exam, similar triangles is cool way to find out.
A very simple method....
Let length of perpendicular be x.
Now by similarity
x^2 = 16 × 9
x = 4 × 3 = 12 feet
16:9 is a very common screen ratio these days.
Here is my method btw with "Inscribed Angle Theorem of Circle"
Suppose 16+9=25 is the diameter of a circle. divide it by 2 and you get the radius 12.5 . Substract it from 16 and you get a smaller triangle of base 3.5 and the hypotenuse of the triangle should be the radius of the circle which is 12.5 . Apply pythagorean theorem and you get the perpendicular length p=√12.5^2-3.5^2 which leaves p= 12. Then the rest you can figure out with the same pythagorean theorem Presh used, 15 and 20. Anyone else who solved in other ways, let me know. Thanks!
I actually used the geometric mean.
Note that the triangle of sides p and 16 is similar to the big triangle (same occurs with the triangle of sides p and 9).
Let b be the base of the big triangle and h its height.
Then:
16/b =b/25
Consequently b is the geometric mean of 16 and 25 so b=√(16*25) = 4*5 =20
With the other triangle you get h=√(9*25) = 3*5 =15
And with any final pythagoras you get p=12
:v
I got it a different way. I set up 3 equations for the 3 unknowns (p, x and y):
9^2 + p^2 = x^2
16^2 + p^2 = y^2
x^2 + y^2 = (16 + 9)^2 = 25^2 = 625
Then I just substituted the values for x^2 and y^2 in the first 2 equations into the same values for the third equation:
(81 + p^2) + (256 + p^2) = 625
2p^2 + 337 = 625
2p^2 = 625 - 337
2p^2 = 288
p^2 = 144
p = 12
Then plug that value of p into the top two equations to solve for x and y:
81 + 12^2 = x^2
225 = x^2
x = 15
256 + 12^2 = y^2
400 = y^2
y = 20
I did the same. no need for the 3-4-5 triangle.
I've done same as this too...
Don't need to know the trick triangle ratio stuff the speaker mentioned, just simple algebra and pythagoras theorm.
It can be solved in several different ways, but if you recognize both triangles as 3-4-5 triangles you can solve it in your head in a matter of seconds.
Anzor Qunash, That's odd, because I don't remember ever applying to MIT.
Awesome. I've never been to Boston. Now I'm gonna live there. Do I need to time travel back to 1869 or can I attend in 2017?
I did this both ways. I solved for P^2 using Pythagorus with the two triangles. More fun... But "intuitively" knowing (ok-- memorizing) Euclids axiom on similar triangles will get you through a test quicker.
Or using the much simpler way....
p = the geometric mean of 9 and 16, p= sqrt(9*16) = 12
Using the 3,4,5 right triangle property.. Since the short leg of the small triangle is 9, and the long leg is 12, then the hypotenuse is 15.
Using the 3,4,5 right triangle property.. Since the short leg of the big triangle is 12, and the long leg is 16, then the hypotenuse is 20
I did it in the same way
I used this method
Yh
It's called the geometric mean theorem... jesus, mindyourdecisions loooves to use the longest solution physically possible to boost runtime.
Aaron Condron
That theorem is based on similarity/Phytagorean therorem.
And, if the hypotenuse is 25 (and that is the only thing that you know), there is infinite number of irrational lengths of the legs.
You can't say "25=5*5 therefore the legs are 3*5 and 4*5".
I couldn't even find the hippopotamus !
its hiding just behind the hippocampus :)
+fridgemagnet
They became extinct...got stuck in the frogmire.
***** stop talking about my mum
I solved this using Euclid’s Theorems and Pythagorean Theorem, i’m the only one?
I think I did what you did: Three triangles, three unknowns (the perpendicular and two legs) and, using Pythagorean Theorem, I wrote three equations. Substitute..substitute and solve for the three unknown lengths. Got the answers shown in the video.
Same. Three unknowns and three equations. Simple substitution.
Yes. That works. But in this case you have a right triangle with a hypotenuse of 25... easy-peasy
Yeah- Wrote the three equations, then subtracted them from each other and then solved for the legs and perpendicular.
@@DanielinLaTuna Yeah, just choose from the infinite number of them
It is also possible to solve with the Pythagorean formula. The vertical side of the little blue triangle is "a" and the horizontal side of the big green triangle is "b". There is also a great triangle containing both of them. Pythagorean formula for the largest triangle: a squared + b squared = 625 ((9 + 16) squared). The Pythagorean formula for the green triangle: a squared = p squared + 81. The Pythagorean formula for the green square: b squared = p squared + 225. Substituting the small triangles to the largest we get: p squared + 81 + p squared + 225 = 625 => p = 12. Substituting "p" into the green and blue triangle formulas we get the length a and b.
I have figured it out, but I had completely missed that the triangles are similar, AND scaled-up versions of the 3-4-5 triangle, so I repetitively used Pythagoras's theorem AND equations, which resulted in a much more convoluted process. This is nothing new for me, my teachers often said I missed the simplest solution and threw myself at the hardest. Here's the complete procedure I've followed:
sqrt(9^2 + x^2) = a
sqrt(16^2 + x^2) = b
sqrt(81 + x^2) = a
sqrt(256 + x^2) = b
a^2 + b^2 = (9+16)^2 = 625
81 + x^2 = a^2
x^2 = a^2 - 81
256 + x^2 = b^2
x^2 = b^2 - 256
a^2 - 81 = b^2 - 256
a^2 = b^2 -256 +81
a^2 = b^2 -175
a^2 = 625 - b^2
b^2 -175 = 625 - b^2
b^2 *2 -175 = 625
b^2 *2 = 625+175
b^2 *2 = 625+175
b^2 *2 = 800
b^2 = 400
b = sqrt(400)
**********
* b = 20 *
**********
256 + x^2 = 400
x^2 = 400-256
x^2 = 144
********
* x=12 *
********
9^2 + 12^2 = a^2
81 + 144 = a^2
a^2 = 225
********
* a=15 *
********
just 21h ago, i used the same way
My answer was 26 and 30. Sorry mom i did not make it to MIT but i saved a whole bunch of money by switching to Geico 😁
Tariq Khan,
😂😂
I'm glad I wasn't eating or drinking something when I read this comment of yours! It would have ended up on the floor! Brilliant reply!
I learned this in 7th grade.
And I'm not even Asian.
*not even Asian* lol
just because you know Pythagorean Theorem doesn't mean you know everything about this problem lmao
@Ethan Crowley Well isn't it good "racism" if you attach something cool to a race... I get pretty happy when that tho.
Noctis what did you mean by saying you are not Asian ?
@@AlthafShameelPP he means that he is not an Asian
Used 3 equations with 3 unknowns, p, a and b. Worked just as well if a bit less elegant.
You could've found the perpendicular by finding the geometric mean of 9 and 16.
You could've found the shorter leg by finding the geometric mean of 9 and (9+16).
You could've found the longer leg by finding the geometric mean of 16 and (16+9).
+R.Q Games Yo,
You confused geometric and quadratic mean. Geometric mean would be sqrt(9 x 16).
Quadratic mean is sqrt(9^2 + 16^2).
Good news everyone! Most of us have been admitted to MIT 150 years ago!
Thank you for posting. I use right angle trig at least 5 times a week as a toolmaker.
I love math and numbers.
Please post more.
I subscribed and liked.
Peace.
Use theorem of geometric mean for the altitude
P Square = 9 x 16
= 144
P = 12 taking square roots
And find sides using Pythagoras theorem
Yeah well knowing how to solve this doesn't make it any easier to find a good chicken burrito in my neighborhood.
unless you perpendicularly square root your location to the vertex of the hipotenuse of Pythagoras restaurants in your area.
Hipotenuse? Is that the hypotenuse of a triangle made with hippos? Golly. I didn't know we humans had that kind of tech.
Hey, this can easily be done by making pair of quadratic equations in 2 variables and getting one arm . Then solving for rest two would be very easy . The equations are as easy to solve as it is for solving pair of linear equations in 2 variables .
I just broke this down into three equations with 3 variables. Solved for P. Easy. The way he solves is faster.
i did the same thing
MisterBinx
you're an idiot.. didn't you recognize the 3/4/5 triangle immediately? you do not to formulate anything
xcarflyx Its not a 3/4/5 triangle though, P = 12 and the legs are 15 and 20
nah, he's right. there's actually 2 3/4/5 triangles, the largest triangle (its sides are 25, 15, and 16, divided by 5 are 3,4,5) and the smaller triangle created by the perpendicular (who's sides are 9, 12, 15, divided by 3 being 3,4,5). I think that the question is also supposed to be a grammar question. Most students should always correctly give the length of the perpendicular, but I'm certain the wording of the question is meant to trap you into giving the lengths of the legs of the smaller triangle, when it's really asking (i believe, correct me if i'm wrong) for the legs of the largest.
ahh fair enough
Notation
Perpendicular of biggest triangle=a
Base of biggest triangle=b
By Pythagoras theorem solve for 3 triangle we will get
25²=a²+b² ...............eq.1
a²=9²+p² ...............eq.2
b²=16²+p² ...............eq.3
By the help of eq.1 and by adding eq.2 and eq.3 we get
9²+p²+16²+p² =a²+b²=25²
81+256+2p²=625
2p²=625-337=288
p²=288/2=144
p=√144=12
So perpendicular
p=12
You skip the proof that triangles are similar!
yea, they all have the same angles.
they have 1 common angle and side, and 1 angle is 90. and thats proven
shitttt if only getting into MIT was still this fucking easy. this like a bonus question from my math tests back in 9th grade
and this is why i took a Trucking job
You can also use the euclid'theorem: p equals to h, the height of the triangle. So, for the second eulcid's theorem it's:
Proiecton of a on ipotenuse(16):h=h:proiection of b on ipotenuse(9)
Actually you get the same results, with the same passages, but you arrive to that point in an easier way.
(sorry for any English mistakes, I'm still learning😂😂)
We can solve this with the help of similarity rule of triangles in seconds!!
a, b - legs of big triangle, c - hypotenuse. 16/a = a/25, a^2=25*16, a = 5*4 = 20, Pythagoras says: b^2=(9+16)^2-20^2=5*45=225, b=15.
@Harsh Sinyal can you explain the a.b=I^2?
finally something I actually could solve!
Couldn't agree more with that assessment
I'm surprised that it was even used. No challenge.
"I'm surprised that it was even used. No challenge."
Little challenge _now_, but consider when this was introduced. No calculators to do square roots and squares. You would have to use a slide rule. And since slide rules aren't exactly the most accurate things (depending on the values involved), I'd bet that they were specifically looking for applicants who were clever enough to notice that the fact that the two triangles were similar provided a much easier (and more accurate) solution that didn't involve going through the Pythagorean theorem and all those extra steps.
That kind of thinking is still incredibly valuable today, though you see it more often in computer science, where a clever solution to a math problem that provides the answer while going through as few computationally intensive steps as possible can be highly valued. For example, look at Stein's algorithm: en.wikipedia.org/wiki/Binary_GCD_algorithm
You wouldn't need a slide rule, the hardest square root you needed to calculate was 225 (15). 144 should be well known from memorizing multiplication tables and 400 is easy.
chinareds54
I didn't say you'd *need* it. I said it'd be less convenient than it is now.
And regardless, my point still stands.
I literally learned this in 8th-grade geometry, wow. Education has drastically changed
Got the answer in my head in well under 1 minute.
Used the rudimentary 3-4-5 right triangle theorem of Pythagoras.
No messing around with square roots, cross multiplying, etc.
Also, applying this simple math, perfect square corners can be made.
Been using this trick for 45 years without needing a speed square.
Omg
This is MIT??
I learnt this in 10th grade 😂😂😂
Prathik Koundinya no it's not, it was a question on the admission exam of 1869, so a question for high schoolers 148 years ago.
I guess it is not the solution, but the method used that separates us from the geniuses
quite pointless. solving a problem in a harder way doesn't make you a genius.
+Turgul I think the point is that a genius finds the most simple way to solve the problem...
yeah i learnt it in my moms womb
Are you serious, my future 6 year old son could solve this shit
I mean, it was 150 years ago
you sure about that? My future 2 year old could outsmart your future 6 year old if it would take your future 6 year old until they are 6 to do this. :P
No. He couldn't.
Why does everyone like the dick reply, a "that was easy, give me harder problems" wouldve suffice
Ansh, these videos are for General Audience, so most of youtube viewers can try to solve the problems at intermediate difficulty.
How come was I able to do it in 1869 BC?
Using the pythagorean triple you can solve this in a sec.
If we have the triple 3^(2)+4^(2)=5^(2) we can multiply every number by five and again obtain another triple: 15^(2)+20^(2)=25^(2) that gives us the length of the legs of the triangle.
I feel pretty good that I got it before looking at the answer.
EDIT: I watched the video, but I solved it differently. I just added 9 and 16 to get 25 as the hypotenuse of the big triangle and treated it as a 3-4-5 triangle scaled up 5 times on each leg. Then I used the Pythagoras theorem again to find the perpendicular.
thats how I thought about it too ;)
i solved it with only applying pythagoran theorem
Yuanxin Liu it ended up right, but you could not assume it is a 3 4 5 triangle scaled up just because the hyphotenuse is 25. There are infinite different combinations of side lengths which would result in the hyphotenuses being 25.
the fact the perpendicular line cuts it into 3 and 4 (9 & 16) probably means it has to be a 3,4,5
Leon Carpenter, was wondering that myself. I sense such intuition is frowned upon here unfortunately.
l instantly saw they were both 3 4 5 triangles
exactly, this was a little easy for MIT
Johann Sebastian Bach Dude it was 1869.
It wouldn't make any difference if it was 1869 or now, they weren't dumber back then. They still knew the same math as we do now, the only advantage we have is the calculator, which you don't really need for this.
They must have had a different system when it came to learning math.
I agreed :)
+Link the thing is, students overall were dumber, not in a poor genes kind of way but more like at that time there were far fewer schools teaching higher level math, thus far more teachers did not have the skill set to prepare students for this. In the mid to late 1800's mostly only children of affluent families were schooled much beyond basic arithmetic, writing, and maybe reading.
Long story short an acceptance exam back then wasn't so much to see who was smart enough to handle a question like that, because it is very easy for someone halfway versed in geometry, but it was more to see if the person even possessed the requisite skills to handle higher math at all. If a person really wanted to go to MIT but didn't even know what 10x20 was, let alone square roots, geometry, trigonometry, calculus, etc. then they would likely fail quickly and therefore have no reason to be accepted into the school.
This is genuinely easier than the 5th grade triangle you showed in another video.
I would have figured this out 15-20 years ago, but not today. The knowledge was tucked away too deeply.
To think that 150 years ago, the average secondary school problem of today was equivalent to the contemporary university entrance exam-level problem.
i learned this in 8th grade?
the answer is yes.
davis kampschror we'll learn this shet next year in the 9th grade...germans don't do this until then
davis kampschror same
davis kampschror I learn the pythagora theorem in 6th grade and all class learn at the same time sow could be betwen 6-8grade
In The Netherlands students will hear this at the First year middle school. Called "Stelling van Pythagoras" In English It's called Pythagoras theorem.
Or... you can just use a system of equations:
Assuming a is the vertical leg, and b the horizontal:
9 ^ 2 + p ^ 2 = a ^ 2
16 ^ 2 + p ^ 2 = b ^ 2
a ^ 2 + b ^ 2 = (9 + 16) ^ 2
Solve the system and you get:
p = 12
a = 15
b = 20
This problem was not hard at all.
It was easy. Just putting x as the perpendicular, doing a system of equations with the two hypotenuses of the internal triangles and solving them did the trick.
First hypo was square of 16^2 + x^2
Second one was square of 9^2 + x^2
So you just put 256+81+2x^2 = 625 and you find x
Then you do 2 pitagora's and you are done
Another way is to use SSS similarity theorem:
9:x = x:16 (evaluate mean and extreme)
x² = 144 (take the square root on both)
x = 12
Take the hypotenuse of both triangles (which are the legs) using Gouga Theorem:
Triangle 1:
√[(9)²+(12)²] = y
√(81+144) = y
√225 = y
15 = y
Triangle 2:
√[(16)²+(12)²] = z
√(256+144) = z
√400 = z
20 = y
Therefore, the altiude to the hypotenuse is 12, and the two legs are 15 & 20.
Hope this helps.
MIT substituted this Exam Questionnaire with : Explain The Big Bang Theory In 8 Dimensions ? Explain The Universe 10 to Minus 94 Seconds old and the mind of God . 500 words or less !!!
42.
1:16
instead of going through all that confusing working, why not just use the euclid's theorem to determine p² = 9×16
that's exactly what he is doing
These are really useful formulas which you could have used to solve this and also is used for lots of others it works in right angle triangle when height is on hypotenuses a' is 16 a is 20 b' is 9 15 is b in this example
ch=ab
a²=a'c
b²=b'c
3:4:5 triangle
Or am I missing something?
That's how I did it.
That's the "intuitive" way to do it. I did have to satisfy myself afterwards that there was no alternative solution. This would have been trickier if the 9/16/(25) didn't tickle the back of your brain that says 3:4:5.
The 3:4:5 rule doesn't apply all the time in this particular situation.
It applies every time in this particular situation.
Rip Sumrall No, if you change the other two angles of the right triangle, the 3 4 5 rule won't apply.
My feeling of superiority is quickly squashed when I watch his other videos
well that made me feel good about my math skills. though i didn't even try the similarity, at all.
instead of that method, I used circles such that the hypotenuse of the overall triangle was the diameter. halving the diameter gave 12.5 = radius. now drawing a new line (the radius) from the centre of the circle to the lower left point forms a new right angle triangle. using pythag: 12.5^2 - (12.5-9)^2 = 144. so, square rooting gives the length of the perpendicular = 12.
Why this complicated explanation? I used the Höhensatz (it's german, I don't know the english one) and two times the phytagoras.
Höhensatz (still don't know it in english): root out of 9x16 = 12
Pythagoras top triangle: root out of 9²+12² = 15
Pythagoras bot triangle: root out of 16²+12² = 20
It's so easy, I learned that in grade 9, Realschule (sth like middle school).
Hallo, was ist los? The Höhensatz is the geometric mean where you take SQRT[a * b], whereas the arithmetic mean is (a + b + c +...)/(number of terms). Viel gluck Englisch lernen! P.S. your English is really good!
Thanks for your answer!
The exact translation of "Höhensatz" would be "altitude theorem [for right-angled triangles]"
Thank you! ^^
It's really the same way. The video just basically incorporates a proof of the altitude theorem instead of relying on the viewer already knowing it as a separate, explicitly stated theorem.
If the perpendicular height is p and one of the angles theta. Then tan(theta)=p/16 and cot(theta) = p/9. product of the two is unity from which we can find p and with subsequent Pythagoras theorem the respective sides. Quite lame for MIT admissions.
Actually it is "quite lame" to need to bring trig into a problem of this simplicity.
That didn't seem thaaaat hard. ._. Typical highschool problem.
You only need a highschool education get into a university. Also check the year: 1869.
Actually the Pythagorean theorem was invented in 500 BC
I used a three equations with three unknowns mechanism:
There are three rectangular triangles, for each of which you can make their own a^2=b^2+c^2 equations.
each of the three have either a, b or c filled in with 9, 16, or 25.
Using x, y, z for the unknown sides you can work out x, y and z
huh?could you repeat the question?
I solved it instantly from the thumbnail alone!
maybe MIT will take you ;)
I got my math degree from duke.
+Pika250 damn! thats cool!
Well to be fair, the thumbnail does have all the required information, as long as you assume where the right angles are.
Big assumption though. They might've just done that to mess with you. I've had trig problems that intentionally did that. >.
I did the same
I solved the problem by fast forwarding the video. MIT here I come!
You could also solve it by using Pythagorean Triples. One triple is 9,12,15, and since it’s a right triangle, and since we’re given 9, we can say that p is the 12 and the side opposite of the right angle (the hypotenuse) is 15, when worked out using the Pythagorean Theorem, both sides add up to 225. So the leftmost leg of the larger triangle is 15, and the perpendicular is 12. Knowing that p = 12, you can then use the Pythagorean theorem again, and do 12^2+16^2=c^2 to find the bottom leg, and you would get 144+256=c^2, which would result in c=20.
You do not know it would be 9-12-15 just by giving 9. There’s is infinitely many triples with 9, one example is 8-9-sqrt(145)
I dare to say that the top is 53° hence making the bot side 20 and the left side 15 let's see if my guess was right?
Edit: i was totally right by using the infamous 53°... Remember guys don't you ever forget about 53° and 37° those are some special snowflakes
@salelltd still an approximation. So not "to be exact" :/
Umm... if the original right triangle has a total hypotenuse of 25, we know (because 3-4-5 triangles) that a and b are 15 and 20, respectively. Then using d^2 +9^2=a^2 (if d is the perpendicular) we find that d is 12. This is an easy geometry problem.
Yes, but a right triangle with a hypotenuse of 25 does not have to have legs of 15 and 20. They could be 7 and 24, or some irrational numbers.
KC3141 it can't be 7 as 7 < 9 and 7 < 16
I'm in 7th grade and this is easy. Just by looking at the thumbnail (assuming I have to calculate the missing sides) I know that the first side is 15, the dividing line is 12, and the last side is 20.
Yup
FunnyMau5 .
FunnyMau5 wow can you be my friend
U can use the fact that the bigger triangle is similar to smaller inscribed triangle. And hence equate the ratios of their hypotenuses and altitudes.
The altitude of the bigger triangle is the hypotenuse of the smaller inscribed triangle = sqrt(81 + p²)
Applying ratios
25/(sqrt(81+p²)) = sqrt(81+p²)/9
225 = 81 + p²
p² = 144
p = 12