A nice Olympiad qualifying question from the Philippines

If a semicircle and its flipped copy are placed side by side, how much distance do they span from end to end? This is a neat problem adapted from a Mathematical Olympiad Qualifying Test in the Philippines. Can you figure it out?
Subscribe: kzbin.info...
Send me suggestions by email (address at end of many videos). I may not reply but I do consider all ideas!
If you buy from the links below I may receive a commission for sales. (As an Amazon Associate I earn from qualifying purchases.) This has no effect on the price for you.
Book ratings are from June 2021.
My Books (worldwide links)
mindyourdecisions.com/blog/my...
My Books (US links)
Mind Your Decisions: Five Book Compilation
amzn.to/2pbJ4wR
A collection of 5 books:
"The Joy of Game Theory" rated 4.2/5 stars on 200 reviews
amzn.to/1uQvA20
"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" rated 4/5 stars on 17 reviews
amzn.to/1o3FaAg
"40 Paradoxes in Logic, Probability, and Game Theory" rated 4.1/5 stars on 30 reviews
amzn.to/1LOCI4U
"The Best Mental Math Tricks" rated 4.2/5 stars on 57 reviews
amzn.to/18maAdo
"Multiply Numbers By Drawing Lines" rated 4.1/5 stars on 23 reviews
amzn.to/XRm7M4
Mind Your Puzzles: Collection Of Volumes 1 To 3
amzn.to/2mMdrJr
A collection of 3 books:
"Math Puzzles Volume 1" rated 4.4/5 stars on 75 reviews
amzn.to/1GhUUSH
"Math Puzzles Volume 2" rated 4.3/5 stars on 21 reviews
amzn.to/1NKbyCs
"Math Puzzles Volume 3" rated 4.3/5 stars on 17 reviews
amzn.to/1NKbGlp
2017 Shorty Awards Nominee. Mind Your Decisions was nominated in the STEM category (Science, Technology, Engineering, and Math) along with eventual winner Bill Nye; finalists Adam Savage, Dr. Sandra Lee, Simone Giertz, Tim Peake, Unbox Therapy; and other nominees Elon Musk, Gizmoslip, Hope Jahren, Life Noggin, and Nerdwriter.
My Blog
mindyourdecisions.com/blog/
Twitter
/ preshtalwalkar
Merch
teespring.com/stores/mind-you...
Patreon
/ mindyourdecisions
Press
mindyourdecisions.com/blog/press

[@volodymyrgandzhuk361]

√2+√6=4 sin 75°

[@MindYourDecisions]

I was intrigued at this answer, so I tried to work this method out myself! Perhaps there is a better way, but here is what I thought. At 3:26 we have a right triangle with legs of 1 and 2 + sqrt(3). The angle at point A thus has tan(A) = 2 + sqrt(3). I didn't recognize it at first, but this is actually a special value for A = 75°. Then the other angle at D is 15°, so we want to find 1/sin(15°) = 4/(sqrt(6) - sqrt(2)) = sqrt(2) + sqrt(6). Pretty neat way to solve using trigonometry! Exact value of tangent of 75 degrees (uses sum of angles and values of 30° and 45°) www.mathway.com/popular-problems/Trigonometry/308167 Exact value of sine of 15 degrees (uses sum of angles and values of 30° and 45°) www.mathway.com/popular-problems/Trigonometry/301128

[@camembertdalembert6323]

@@MindYourDecisions 75° = 30°+45°. Then you can evaluate sin(75°) with the formula sin(a+b)=cos(a)sin(b)+sin(a)cos(b).

[@AliKhanMaths]

That's a great method of solving it!

[@pavelkotsev1542]

@@MindYourDecisions impressive! :) thank you very much! Do you consider a follow up video? Seems to be a pretty interesting thing that those angles turn out to be so nice.

[@volodymyrgandzhuk361]

@@camembertdalembert6323 you meant cos(a)sin(b)+sin(a)cos(b)

[@low-litlight3438]

Honestly surprised by how relatively simple this is.

[@Superman37891]

Same

[@manamimnm]

remember that it's for 15 years old kids to solve

[@Superman37891]

@@manamimnm ok but 15-year old Olympiad-level mathletes should probably also be able to do this in their heads.

[@low-litlight3438]

@@manamimnm I mean, I know of problems for the same purpose that are much harder than this. Not saying it wouldn’t be very challenging for someone that age, but it does seem too easy for an Olympiad Qualifier.

[@adityajha5500]

@@Superman37891 yes

[@caspermadlener4191]

Olympiad questions are special. They never require some advanced mathematics, but they are still able to let you think in a creative way, because every problem is different.

[@batchrocketproject4720]

I got as far as √(8+4√3) in my head but had no clue how to resolve that expression further. The progression of second half of the video was really pleasing to watch. Thanks.

[@Zorlig]

Me too

[@kurtiskuchcinski2628]

Same. Got stuck at that point.

[@antoniopedrofalcaolopesmor6095]

Me too

[@akashraj6391]

kzbin.info/www/bejne/ooG8qYxvZdOKbq8

[@sakshitandel8572]

8th standard mathematics

[@sinaydos4754]

Damn after being a subscriber for two years I finally managed to solve a problem by my self; im proud of myself

[@syedmohdjaved]

👍

[@youtubeaccount1718]

Bro its ez

[@akashraj6391]

kzbin.info/www/bejne/ooG8qYxvZdOKbq8

[@AliKhanMaths]

Love that first feeling of satisfaction!

[@AnshuOP69]

Same

[@N2O_The1000thElement]

When I was in middle school, we students weren’t allowed to draw anymore lines into the drawings because it would make it easier to solve it and it wouldn’t require specific algorithms anymore but now that I know this is how math problems should be solved, I’m now going practice doing this. This channel has opened my mind in math and showed me how I should approach math. +1 subscriber

[@theairaccumulator7144]

How did your teacher think those algorithms were discovered? That's some low iq reasoning.

[@Propane_Acccessories]

This is why my kid will not be taught math in public school lol

[@aloosh1375]

@@Propane_Acccessories Doesn't mean that private is better

[@Propane_Acccessories]

@@aloosh1375 Not always of course. It very much depends on the parents as well.

[@espomaths]

I solved it as follows: I drew a line from D to the intersection point (called M), then the perpendicular from M to the lower line and called their intersection point K. Then I considered the two right triangles CMK and DMK, and exploited two facts: 1. CK + DK = CD = 2; 2. DK : KM = KM : CK (by the proportion of right triangles with one side in common). From 1) and 2) one can find CK, DK and, by Pythagora's theorem, DM. The result we're looking for is 2DM - the double root trick will give the desired form. The solution as Presh presented is more elegant, but this is another way nonetheless!

[@abh-1-shek]

Wow! Presh's accent makes it legit easier to understand. Wish he taught me maths. Didn't know Olympiad had easy questions too.

[@Jkmthink]

Yeah The easy ones remain hidden like treasures

[@akashraj6391]

kzbin.info/www/bejne/ooG8qYxvZdOKbq8

[@user-xb8zo6wc9h]

Yup, not difficult. Just more steps and ppl give up ☺️

[@glennsampson5945]

When solving the two equations at the end, it is easier to take advantage of the fact that x and y are positive integers. The equations are xy = 12 and x + y = 8 There are only 3 possibilities for the first equation, 1-12, 2-6, 3-4. The only pair that sums to 8 is 2-6. Hence, x =2 and y = 6.

[@rohangeorge712]

ikr but same thing practically

[@touristguy87]

...but it's a lot more fun to use complex math!

[@piman9280]

I love the way Presh explains things, especially when he avoids those "weird" names for standard formulae. Great video.

[@akashraj6391]

kzbin.info/www/bejne/ooG8qYxvZdOKbq8

[@kvdrr]

@@akashraj6391 Damn, it's truly saddening to see what lack of toilets does to a man. I'm truly sorry for you and your horrible accent.

[@draganandrei5356]

Gougou theorem avoided! Near miss.

[@JamesWylde]

Except that he should just say Pythagoras because that's what's EVERYONE knows it as, and he just obfuscates the problem.

[@piman9280]

@@JamesWylde - just be thankful for small mercies. LOL

[@hasibulislamshanto143]

I solved many problems of your channel. But I still feel so proud. Thank you.

[@sohamkulkarni5674]

Consider the intersection pt. as 'O'. The arc BO subtends angle 30° (sin(1/2)) at the centre of the left circle. Therefore, angle BAO=15° Now, AD.sin (15°)= 1 AD= √6 + √2

[@JLvatron]

I didn't get the radical conversion before. I tried it again months later and I solved it. Thank you!

[@euromicelli5970]

5:45 it’s a small refinement and saving, but we’re already done right there. The system of equations is symmetrical to swapping x and y, so the quadratic will necessarily have as solution the values of x and y.

[@curtmack]

In the process of factoring x^2 - 8x + 12, you had to answer the question "what are two numbers such that their sum is 8 and their product is 12", which is just the question you started with in "solve 8=x+y and 12=xy". Rewriting it as a quadratic does allow you to solve using the quadratic formula, in case you can't find it in your head.

[@TheQEDRoom]

Exactly what I'm thinking. I feel it is similar to when we are kids we have 18-9 where you need to regroup since 8 is bigger than 9. And so we ended up with the same exact problem. Then when in high school we have that. :)

[@jannegrey593]

Yeah - you're correct - it is an unnecessary step - that being said it might be useful to know how to solve this type of questions (this channel wants to educate people) and we don't know if for example you were not given points for some proper step by step explanation - depends on a test I suppose. This was quite easy question for this channel (so something I could have resolved on my own - without much trouble) and people going for Olympiad often trip on improper explanation of the answer. Depending on what level you are obviously.

[@seroujghazarian6343]

Well, you could use Discriminant=64-48=16 x=(8(+ or -)4)/2=6 or 2

[@mcmac8027]

I didn't even care to read the title until I started watching the video and shocked to see it was from the Philippines. Amazing!

[@smellafella2841]

I love that you can tell that math makes him so excited in every video

[@nzeches]

Few suggestions for simplification : 1. By symmetry it is quite obvious that both the line joining both centers, and AD are going through the tangent point which is center of symmetry ; from there we have 2 right triangles to calculate. 2. Once you reach xy=12, there are not so many possibilities with x,y integers - checking for x=1,2,3 is sufficient to find the single solution that also satisfies x+y=8 where x

[@pronabchanda769]

The process of solving the equation is really great. I like it very much.And also learnt a new process of equation solving without calculator.

[@yippee8570]

I knew what needed to be done, but I did not have the mathematical knowledge to solve it. I'm not sure I do now, but I was able to follow (most of) it. Great explanation and a beautiful puzzle!

[@eannacoleman957]

I'm so pleased that I was actually able to do this one myself before the video

[@paulrubens776]

what a great problem, i liked how it all unwrapped to the solution

[@musicmaniaclover]

Having established the length of the first triangle’s long leg (square root of 3 or 1,7320) we have for the second triangle: Short leg = 1 Long leg = 1 + 1,7320 + 1 = 3,7320 Therefore the distance A to D (the hypotenuse of this right triangle) = 3,86 Simpler

[@AndRB859]

I agree. The distance between the semi-circles’ centres is 2x(sqrt(1-0.5^2))=sqrt(3). Hence horizontal distance AD is 2+sqrt(3) and vertical distance is 1. Much easier than the supplied method (for a change).

[@bienvenidos9360]

Seriously. What was the need to solve for x and y? The question was to solve for the distance of AD.

[@satyapalsingh4429]

Very nice problem .Explained beautifully .Thanks ,dear professor

[@rajmankad2949]

What an elegant solution

[@gopanneyyar9379]

In the last part, there is an easier way. (x-y)^2 = (x+y)^2 - 4xy so that, you just need to solve simultaneous equations x+y= 8, x-y=4

[@Bluhbear]

That would give you x = 6, y = 2, but the answer you want is the other way around.

[@alirezalashkarboluki7703]

It's definitely AMAZING!!!

[@veli-mattijantunen3803]

Thanks for the nice solution. You could also use the power of the point approach. First calculate the power for the point of the left circle center, thus (1+y)(1+y)=1·3, where y is the distance between point B and right circle tangent (T), then using this value (y=√3-1) you can solve the power for the point A, (AD·AD/2 = T·T) i.e. ½x·x = (2+y)(2+y) getting the value for x as √2+√6.

[@donmichie1193]

Hor length of small triangle = 1.732 (sq rt 3) Hor length of large triangle = 3.732 Thus AD = sq rt ((3.732 x 3.732) + (1 x 1)) Thus AD = sq rt 14.92782 = 3.86

[@netboy288]

lovely to see there are so many different approaches! I used the cosine rule to find the distance between D and the point where the semicircles meet. (by subtracting (from pi) the bottom right angle of the central triangle which you also constructed, which ends up nicely being pi/6)

[@AvoidsPikes-]

It's faster.

[@nporob9900]

U deserve 50 million subscriber .Love from🇧🇩🇧🇩🇧🇩

[@TheQEDRoom]

Nice one. But I always find the last step for this a little funny. Find two numbers that adds to 8 and are factors of 12. And we use substitution to make a quadratic equation then use factoring (which is basically finding two numbers that add to -8 and are factors of 12) :)

[@cannot-handle-handles]

Haha, good point!

[@AceInAcademy]

Great work " Presh Talwarkar " your work is very inspiring and interesting to watch, keep up the good work!

[@stromboli183]

“Our favorite right triangle theorem”, that’s a subtle way to put it 😄

[@SM-gh3cy]

I used another way to solve the problem using similar triangles. It was fun.

[@Terminarch]

This was so unbelievably more difficult than getting the literal answer, needing it in a certain form. sin-1(.5) = 30 degrees (because they must intersect halfway the vertical height) cos(30) = x/.5 ==> x = .86 (this is the horizontal position of intersect from one circle center) add 1 for the other radius half we skipped, 1.86 is the horizontal distance from one circle outer edge to intersect double for 3.73, total horizontal distance (not surprisingly slightly under 4) 3.73^2 + 1^2 = h^2 h = 3.86

[@LiveHappy76]

Excellent solution!

[@anandarunakumar6819]

I did using trig. 4cos(a) is the total length AD, where 'a' is the angle between AD and one of the base line of diameter. Vertical length between two parallel lines turn out to be 2sin(2a), which is 1. Thus angle a is pi/12. Next step is to express 4cos(pi/12) in the required form, which simplifies to sqrt(2) +sqrt(6).

[@trumplostlol3007]

Before you constructure a line through the two centers of the two circle, you need the "symmetry" argument. Without the symmetry argument you can't be sure whether you can draw a line through the two centers AND the tangent point of the two circles.

[@michaelz6555]

"...our favorite right-triangle theorem..." So no specific attribution as opposed to "not Pythagoras." I guess that's progress of a sort :-)

[@Qermaq]

Well "Gougu" was not an attribution, it's the historic Chinese name for the theorem. Not a person's name.

[@spicemasterii6775]

Gougu is now persona non grata

[@mohamedhusam8189]

"the not pythagoras" 😂😂😂 is better

[@JamesWylde]

Presh is making some progress. He's still a pompous ass that fails miserably in his goal of making math approachable and understandable with his resistance to referring to Pythagoras' Theorum the way everyone knows it it as, but I guess he's OK being a modern mathematical laughing stock. Let's face it Presh, half of your views are by folks wanting to see how you contort yourself to avoid saying Pythagoras. People go to 3B1B, Mathologer, and the like to learn something, we watch you for a chuckle.

[@maxlindemann1429]

@@JamesWylde Damn, things really heating up in the KZbin math community

[@InfinityBanana]

I accidentally solve like you and get the correct answer for the first time! Thank god!

[@Mdsamim-go8rp]

How so interesting, so tricky maths problem. Who love this channel😍😍😍❤❤❤

[@grolfe3210]

It seems easier to just work out the values when you get to the smaller triangle at 3.05 and then use actual values to get the longer triangle. I was fine with the diagram bit but lost on the x and y bit!

[@MrEduardo]

I’m happy I’ve made a different solution from you, for the first time. Most of the time I can’t even solve it. I made a right triangle inside a semicircle, with its right angle at the tangent point, and hypotenuse at the semicircle diameter. This triangle has an angle of 15 degrees, and I used its cos, (sqrt(6)+sqrt(2))/4. Great video. As always, mind your decisions!

[@vhm0814]

Same as mine. But in this case, firstly you must prove that the 3 points [A - D - the tangent point] are collinear. It's not too hard but I think that is the most tricky part, many people forget that.

[@rahinbinraiyan2955]

*INCREDIBLE*

[@scorpiodude2751]

Obviously, the 2 semicircles are the same so applying the Pythagorean theorem, horizontal distance pt. 2 to 2 is square root of 3, now assuming that the radius equals to 1, therefore the horizontal distance A-D equals square root of 3 + 2 = 3.732

[@ps.2]

Horizontal distance, yes, but the question asks for the total distance. And that it be expressed as √x+√y with integers x, y.

[@johnmiller5629]

Like most others, getting to the √(8+4√3) solution was fairly simple. I agree that putting it in the required form for the answer was a neat trick. What I'd like to understand is whether there is a good reason for employing such a trick in other circumstances. I suspect the answer is yes, but can't think of any myself.

[@leif1075]

What trick?

[@3057luis]

That´s analytical solutions versus numerical solutions. It looks a bit of a contradition but you can use analytical geometry to compute the numerical solution for most of these kind of problems.

[@ThePieOfTheUnknown]

Idc if its math, I see Philippines I click I just finished watching, its actually quite interesting, I learned things here

[@MaximQuantum]

I got it! I am happy now.

[@SpaceCadet4Jesus]

After the basic premise, I guesstimated 3.80. The explanation was more than I could handle but when it came up 3.86, I knew my old guesstimator was still in operable condition. And as we used to say "Good enough for government work".

[@Exler_Ko]

This question were also in kangaroo math, I thought for so long and not knowing that this were so simple when you showed the solution 😊

[@anandkarthik8833]

My solution was slightly different but after watching the video I see that what I did was essentially the same but conceptually different. I basically imagined that to be in a co-ordinate system, so basically I used the equation of a circles. To deduce the distance in the x-direction between the 2 centres.

[@smoog]

You could use the Cosine rule to find the length from D to the tangent point: C²= A² + B² - 2ABCosC°, where A & B are the radius 1 and C° is 150°. This gives C² =(2 + √3). Times by 2 to get the entire length 2√(2 + √3), or √(8 + √12), which equals 3.86.

[@chintansojitra2102]

great work man , i wonder how you are animating . which softwear do you use for it ? Can you teach us how you make your video ?

[@BytebroUK]

I got to it (eventually!) a slightly different way, but the two methods are equivalent, I think. (Don't have the software, or I'd draw you animated pictures!) Loved this, and thank you for 'scrubbing the rust off my brain'!

[@StepwaveMusic]

I actually solved this problem using the trigonometric identity. At y = 1/2, x = +/- 1/2 * sqrt(3). You can use this to calculate the horizontal length, whereas the vertical length is of course 1 (the radius of the circle).

[@MathSolvingChannel]

this trick is formally referred as de-nesting the radicals, sqrt is the most common case~

[@adveshdarvekar7733]

This question was awesome.

[@no3339]

I tried doing a different method in the first part to find the horizontal length between A and D. The tangent point between circles is denoted E and the tangent point between the right circle and top line is F. The angle EBF is 90. By symmetry of the pseudo-triangle EBF that was created, angle BEF must be twice that of BFE. There values are then 60 and 30. The length from B to F is then cos(60) = sqrt(3)/2. The entire horizontal length, is then 2*(sqrt(3)/2)+2*radius = sqrt(3)+2.

[@supercat4259]

bro that wizardry after the actual problem was the hardest part

[@philip8498]

i had the distance figured out correctly, but that neat bit of shuffling the numbers around to make a nice simply sum of tow square roots is far to advanced for me...

[@pbondin]

An interesting insight into this problem is to realise that, since AB and CD are diameters, then (just focusing on the semicircle AB), the triangle formed by A, B and the tangent point between the two circle is a right triangle. The height of this triangle is 0.5 (halfway between the parallel horizontal lines), and its base is 2. Thus, its area is 0.5. You can ‘stack’ the identical triangle formed by C, D and the same tangent point, on top of the first triangle to get a rectangle. You will observe that the area of this rectangle is, therefore, equal to 1. If you let the longer side of the rectangle = x, the shorter side = 1/x, and we write x^2 + (1/x)^2 = 4 [the square of the diameter]. Solving from there is straight forward, but the manipulation of the surd (radical) expression in the second half of the video was a delight to watch! Thanks Presh.

[@apostolosspanelis5686]

While thy are right triangles, the height is not 0.5 because AB is the hypotenuse and not the base..

[@apostolosspanelis5686]

You really need to apply some trigonometry to find the actual height

[@pbondin]

@@apostolosspanelis5686. Certainly, AB is the hypotenuse of a right triangle, but the area of any triangle is a base (choose any side of a triangle - in this case, I’m using the hypotenuse of the right triangle) x height of the triangle (measured perpendicular to the chosen base) x 1/2. Given that the base that I have chosen runs along the top horizontal line, its perpendicular height measured vertically down is 0.5 (the tangent point is halfway between the two parallel horizontal lines = 0.5. Therefore the area of the right triangle = 1/2( 2 (base) x 0.5 (height) ) = 0.5. Do you follow my reasoning?

[@apostolosspanelis5686]

Yes got it now!

[@Mathematician6124]

I got it right in my first attempt. It does not matter. It matters what a fun it was to solve it. And I am really satisfied with the problem.

[@moisesbarrera4849]

¡Beautiful!. Thank you.

[@Viveksingh_18]

Yeahhhhhh I did it without your solution My maths improved alot because of you Love from india

[@daitedve1984]

We have significant fact that circles are equal, meaning their touch point stays exactly at half of radius = 0.5; Knowing that and radius we can know horizontal distance from center of circle till touch point. Two of such distance + diameter = leg of triangle where we should find hypotenuse. Easy as 1-2-3!

[@rogergriffiths3345]

Made a meal of that one! Mid point between A & B to mid point between C & D has a length of 2. Therefore base of triangle is 1.732. Larger right angle triangle has a base of 3.732 & a height of 1. Therefore its hypotenuse is 3.8654. Done!

[@zoran.grujic]

There is another less complex way to solve the last part. When grouping alike factors xy = 12, with x, y element of N leads to x, y = 1, 12 or 2, 6 or 3, 4. Then from x+y=8 we find it must be x=2 and y=6. The problem and its solution are well presented in this video.

[@royschering1140]

If you define point A at coordinates (0,0), then the coordinates of center of the first semi-circle are (1,-1) and the point tangent to the line above the center are (1,0). A line from this point to the center of the second semi-circle is 2 units long and distance between the two centers is 3^0.5. So the coordinates of the second semi-circle center are (1+3^0.5,-1) and the coordinates of point D are (2+3^0.5,-1). Therefore the line AD is ((2+3^0.5)^2 + (1)^2)^0.5 = (8+4*(3)^0.5)^0.5 = 3.8637...

[@caweddan6004]

Finally I can solve a problem from MindYourDecision

[@cleverarsey3568]

distance between the centre of the 2 semicircle will be 1+1 = 2 so the horizontal distance between the 2 centres will be sqrt(2^2 - 1^2) then find horizontal distance of AD is above answer + 1 + 1 because of 2 radiuses then final answer is just sqrt(1^2 + horizontal distance) = 3.86

[@SogeYann]

Find in using trigonometry. Angle BAD = 15° then AD = 2×2×cos pi/12 = (sqrt2)(sqrt(3) +1)

[@sgcomputacion]

Before watching the video AD = √2 + √6. Thanks for making my brain work!

[@smurphsFTW]

Finally one i figured out on my own!

[@user-er8kg3cx8v]

The circles are tangent at the point where sin = 0.5 as radius = 1. Take radians for sin(X)=0.5 and apply to cosine. Cos(30) = 0.86 roughly. Take length of circles 4 - (1- cos(30)) = 3.72 √1+(3.72)^2 = 3.86

[@logiciananimal]

Nice problem. On the "Pythagoras or not" question - the attribution is to the *Pythagoreans*, if anyone. Standard references on the preSocratic philosophers go through the very tricky question of attributions to the master, if indeed there was one.

[@kermit328]

Simpler solution : The figure is antisymmetrical - the two circles are thus tangent at Y = 0.5. The angle between the X axis and the line leading from the centre of a circle to that point is 30°. Half of AD is the base of the isosceles triangle with sides 1 and a vertex angle of 150°. Therefore AD = 2 x 2 x sin(75°) = 4 sin (30 +45°) = sqrt(2) + sqrt(6).

[@nikhilshenoy1555]

This was my way of doing it: The point of intersection is at the half way point for height meaning that it is 30 degrees because sin(30)=1/2, then cos(30) is (sqrt3)/2, then subtract this from 1 to get half the distance and then multiply by 2. Then take the diameters and add them to get 4 and subtract the value we got before. The height is one since it is a semi circle Then use the Pythagorean theorem to solve for the distance. Didn't know how to do it for x and y

[@kid4411]

At 3:57 , we can also directly get √2 + √6 as the ans , we can eliminate the bigger square root if we get a square term in the inside , thus we can try to look for (a+b)^2 identity directly in 8 + 4√3 , we notice a^2 + b^2 = 8 , 2ab = 4√3 , ab = 2√3 = √ 12 . and 12 = 3x4 = 2x6 , but we know 3+4 = 7 , which isn't equal to 8 , so we know a= 2 and b=6 , therefore we get 8 + 4√3 = ( √2 + √6 ) ^2 . Thus we get our ans . This method is just to simplify things and get a different approach too . Thanks Thanks...

[@camsy83]

I worked out the chord length for one circle at 150degrees and doubled it for two circles, no simultaneous equations required

[@shantanusrivastava9898]

A much easier solution can be found as follows. Drop a perpendicular from A and connect the centres through the point of contact of the two circles. Also drop a perpendicular from the centre of the circle. Using Pythagoras, the horizontal line connecting the centre of the first circle to the point of contact to the other circle is sqrt 3. The horizontal distance between the extreme edes of the circles is 1 + sqrt 3 +1 or 2 + sqrt 3. Then using Pythagoras AD is the square root of ( 1 + ( 2 + sqrt 3 ) squared) . Or the sqrt of (8 + 4 times sqrt 3).

[@karldavis7392]

It was easier than most of them. I did it without a system of two equations, and my square root didn't look as elegant, but I got the same answer in the end.

[@brendanfan3245]

if you check the National college entrance math exam in China （2022 edition 1）, you will find this level question only has a value of 1 or 2 points (out of 150) as a multiple choice. Basically, you have to sort it out in 1 minute.

[@BlackFiresong]

Interesting to see your solution! I started out drawing a similar central triangle to you, but then I deviated in my method. Figured I'd share my solution too as I rather enjoyed it! Instead of working out the missing length of the 2-1-√3 triangle, I worked out that the angle opposite the side of length 1 is 30° (arcsin(1/2)). Based on this, I was able to draw a new triangle between A, the center of circle AB, and the tangent point at which the two circles meet. This is an isosceles triangle with angles 150° (180 - 30), 15° and 15°. The two equal sides have length 1 (as they are both radii of circle AB) and the side opposite the 150° angle is half the length of AD (since you could draw an identical triangle in circle CD and the two line segments passing through the tangent point of the two circles would form the length AD). Therefore, by the sin rule, ½AD / sin 150° = 1 / sin 15°. We know that sin 150° = sin 30° = ½. So ½AD / ½ = 1 / sin 15° => AD = 1 / sin 15°. From here, we just need to work out the value of sin 15°, which is sin(45 - 30) = sin 45 cos 30 - cos 45 sin 30 = (1/√2)(√3/2) - (1/√2)(1/2) = (1/(2√2))(√3 - 1). Multiplying both the numerator and the denominator by √2 gives ¼(√6 - √2). Therefore AD, which = 1 / sin 15° = 4/(√6 - √2). Multiplying both the numerator and the denominator by (√6 + √2) gives 4(√6 + √2)/(6 - 2) = √2 + √6 (i.e. x = 2, y = 6).

[@jack-xf6il]

This is pretty much how I got it, and I thought it seemed more straightforward than that squaring both sides and expanding stuff. But maybe it would have been harder if the 4 didn't cancel itself out at the end...

[@aleksejsivanovs9355]

That 30 degree angle is obvious, because the radius is 1 and the distance between the point of the tangency of both semicircles and CD is 1/2. This allows you to calculate the half of the horizontal leg of the right triangle which has hypotenuse AD. Multiply the leg (sqrt(3)/2 + 1) by 2, square it, add 1 (vertical leg squared), get the same answer.

[@AFSMG]

Lástima tener que verlo subtitulado. Saludos desde España. Excelente trabajo como siempre.

[@minecraftman5679]

it was actually pretty easy as i was able to solve it in my mind but i loved it!!!

[@saadmoquim8624]

Sets up a coordinate axis like a boss

[@StefanMagMathe]

I solved it differently: the semi-circle is given by the equation y = sqrt(1-x^2), so for symmetric reasons we are looking for a positive solution y = 1/2 which leads to x = sqrt(3/4) and the total length (not the diagonal) is 2 + 2*sqrt(3/4). = 2 + sqrt(3). The Pythagoras leads to the result AD = sqrt((2 + sqrt(3))^2 + 1^2) = sqrt(8+4sqrt(3)).

[@emponator]

The point where the 2 circles touch is exactly halfway the two lines. Make a triangle there from the center of the circle, the point in the middle of the 2 and to a line and solve for the length of the line. (sqrt(1^2-0.5^2)) Then make a triangle from point A to the middle and to the same point straight above from the middle and solve the hypotenuse. (sqrt(1.866^2+0.5^2)). Then just times that by 2 and you get 3.86 and change.

[@robg7892]

without the first proof, you know by symmetry that point T must be midway between AB and CD, therefore 1/2. and you know that the radius is 1. same thing though, you just have two little triangles with hypotenuse=1 and one leg =1/2 instead of one big one with hypotenuse = 2 and leg = 1.

[@brandonkay8719]

I solved it in a similar way, but a bit more trig heavy. Both circles are unit circles. The distance between the two lines (top and bottom) is 1, so the height of the tangent point is 1/2 if we consider the height of the bottom line to be 0. Therefore, the horizontal distance from either of the circle centers to the tangent point is cos(30) or sqrt(3)/2, via the unit cirlce. Therefore, the base of the tirangle is sqrt(3)+2. Pythagorean theorem for the rest, and we arrive in the exact same place. Overall, very similar solution, just skipping some geometry and circle properties in favor of a little extra trig.

[@vinayakraj7065]

Hey Presh, I solved it by deducing that the angle formed between two centers and any base line is equal to 30 as the hypotenuse is 2 and perpendicular is 1. This makes the angle BAD equal to 15 degrees as the angle sub tended on the center is twice the one at the circle. And now the distance AD will be twice the distance from A to the contact point which is 2*r*cos15. So total distance is 4rcos15. Since r is 1 and cos 15 is (sqrt3 + 1)/(2*sqrt2) the total distance is sqrt6 + sqrt2.

[@skipmars7979]

You actually had the answer at 3:04 which is how I solved it. your AD y is the radius 1 and AD in x is 1 (radius) + sqrt of 3 + 1 (radius) again. pythagorean theorem to get 3.86. No need to go around your elbow to get to your thumb.

[@jayeshkumar3861]

Enjoyed the morning with question..... I solved this entire question mentally and got correct answer in less than 2 minutes.... 1+ minute gone in thinking. Calculation part was simple.

[@davidnelson9288]

Base is 2 diameters = 4 , Height is 1 radius =1. AD = right triangle hypotenuse SQRT(4^2 + 1^2) = SQRT(17) = 4.123...

[@johnwitha]

Great solution

[@cyrinebenammar2948]

I found that AD is equal to 3,8 using a shorter method : I applied phyta on the same bing triangle with AD the hyp Saying that Ad^2=AG^2+GD^2 Ag=1 Gd= 4-(1/4)

[@touristguy87]

I would say ~4r it's at most r+4r (the long and short side of rectangle AD) so the hypotenuse of that triangle

[@jeanf6295]

For the form change from √(8+4√3) to √6+√2, I took a different approach, For an expression of the form f = √(a+√(b))+√(a-√(b)), if a²-b = c², as it is the case here, then f² = 2(a+c) is an integer. To get the alternative expressions, you just need to use the quadratic formula on the polynomial : P = (X-√(a+√(b)))(X-√(a-√(b))) = X²-fX+c

[@xarran]

Ez. Two circles intersect in the middle, so we know that sinθ=1/2, Thus, θ=30° and cosθ=√3/2, that's the distance from center of a circle to intersection. Add +1 and multiply by x2, so we get AB+CD horizontal distance = √3+2. Use √((√3+2)^2+1) to find AD, which is √(8+4√3), then rewrite as sum √(a+b)^2, which is just (a+b). Then do a little guess work so that a^2+2ab+b^2 = 8+4√3, which is easy using whole numbers.