I googled it but didn't get back your pic, or any AI result for that matter. That was in Firefox. So I decided to switch to Chrome (which I pretty much never use anymore) and indeed I got exactly the same result as your did. How exciting! 😃
@@anghme28ang11Chrome is the browser, acting as your way to the internet. Google is just the search engine that you search web pages. Chrome uses Google as it’s default search engine, but you can change it to something like DuckDuckGo for more privacy, while using chrome as it’s browser.
@@anghme28ang11 'Google' is a corporation which makes a search engine called 'Google Search' (often shortened to 'Google') and 'Google Chrome' which is a web browser (often shortened to 'Chrome'). This is all academic because Firefox and Brave are both better than Chrome anyway.
@@anghme28ang11Crome is Google’s web browser, Google is the search engine run by Google (the company) - so when searching on Firefox (a different web browser), even though they used Google (the search engine), they didn’t get Google AI’s summary, as that seems to be attached to the Crome browser, not the Google website.
Omg this is the first time I’ve actually been able to solve the problem before Andy explained it although I did do it a different way but it’s geometry so it doesn’t matter which way it’s solved
No need for "side splitter theorem". The triangle contained within the orange semicircle is an equilateral triangle in its own right, with side 2 and height root 3.
Then you can draw a line from the midpoint of the top triangle’s base to one of the end points of the bottom triangles base (the length of this = R), this line forms a triangle, you know that the shortest side = 1, and the third side = 2, the angle between these two sides is 120 degrees (180 - the 60 degrees from the top triangle), you can then use the cosine rule to work out R. R = sqrt(1^2 + 2^2 - 2(1x2)(cos(120 degrees))) R = sqrt(5 - 4cos(120)) R = sqrt(5 -(-2)) R = root 7
The question is easy because we directly assumed the midpoint of the smaller equilateral triangle as center, proving it is actually the center and infact the center for both semicircles seemed a lot more challenging.
Euilateral trianle means all sides are 4, so each hashed segment is 2. Split the 🔺️ in ½ and you get 2 30/60/90 Special 🔺️s. Once you appreciate that, the math takes care of itself
After watching this I first searched S.P. Theorem in the incognito mode, no result. Then in normal chrome and got a picture but different and after clicking on that image and scrolling below there was your image after few images. So, i think due to better answer of yours it is either showing on top or in top top 10 image results alongside. Thanks Andy. How exciting!!!🖤
You can easily get the orange radius by knowing that the hypotenuse is 2 and the and the base of the right triangle is half of that since it’s an equal lateral.
I had a different solution to this problem. I proved similarity by angles and used pythagorean theorum on the smaller triangle. Using the height i found the area of the smaller semicircle. Then i used pythagorus to find the entire larger triangles total height and subtracted the radius of the smaller circle. Then i finally used the same method for the area of larger semicircle.
I tried it a different way and got messed up lol. I tried to take the bottom left corner of the triangle, form it into a right angle with the other end going to the bottom midpoint of the circle, and then a straight line up (the hypotenuse) going straight up and bisecting the circle. Top was then 30° so it's a 30/60/90 triangle. I messed it up because i somehow got 27π/4 as the area of the bottom semicircle lol. The worst part is that i also tried to do a normal right angle thing like you did with the 2 and √3, but in my mind i thought it was going to be a 45/45/90 and when it wasn't i just abandoned that idea for some reason lol. Great video :) the side splitter theorem thing shows up on google when you use chrome but not firefox it seems, odd how its browser dependent lol
notice how the triangle inside the orange semicircle must also be equilateral, and notice further that the smaller triangle in the orange semicircle can fit 4 times into the complete triangle. then you can simply pythagoras your way to the solution.
4:26 Wouldn't the probability of that happening just be 1/13? Because even though the dices have different probabilities for each number, the probability of the card matching it would always be the same, and there are 13 options
The minimum value you can get by rolling 2 die is 2 (2 ones), and the max is 12 (2 sixes) So an ace (corresponding to 1) and a king (corresponding to 13) would never match. Taking that into consideration, the 1/13 chance would probably not work
@@kshitijbenedict8431 I know but doesn’t matter, think like this: you roll the dices and it shows a number from 2 to 12, then from the 13 different card options, only 1 of them would match the dices. Hence the 1/13
@@kshitijbenedict8431 It is 1/13! whatever you roll doesn't matter, there is 36 dice combinations and each has a 1/36 chance. Even if you calculate it separately. Each one of those 36 combinations has the exact 1/13 of a match with the card. (i.e. (1/36 * 1/13) *36). You can do it the hard way of getting the probability of each possible number (2,3,4,5,6,7,8,9,10,11,12) but that would be pointless.
I used little r later in the problem, that is why I rooted it. I didn't root Big R since I didn't need it later in the problem. I hope that makes sense.
Why use side splitter? Top triangle is equilateral so you know splitting it in half gives you 30-60-90. Sides are then 1, 2, and root 3. Seems quicker than side splitter theorem (but lacks AI reference to Andy Math).
Ah yes , side splitter theorem....the thing we learnt just 4 months ago. Only that it is for us called "Basic proportionality theorem" or BPT for short.
I'm confused why at 1:43 the side can be split from 4 into 2 and 2. What guarantees that the side split line between the half circles is splitting the side exactly in half?
A pair of dice do not have an equal probability to produce a number between 2 and 12. To roll a 2, you can only have 1-1, just as for a 12, 6-6. But for a three, you can have 1-2 or 2-1, for an eleven 5-6 or 6-5, and so on up to rolling a 7 (1-6, 2-5, 3-4, 4-3, 5-2, 6-1). So, the odds of rolling a 7 and picking a 7 are (1/6)(1/13), or 1/78. But the odds of rolling a 2 and picking a 2 are (1/36)(1/13), or 1/468, the same as rolling a 12 and picking a queen. You can easily calculate the other potential outcomes simply by looking up the probability of rolling a number with 2 dice and multiplying it by the common (1/13) for then pulling the right card. I suppose one trick could be that you have a 100% probability of getting a number between 2 and 12 with the dice, and therefore the 1/13 for then pulling a card that matches the specific number, but that feels like a cop-out.
@keith6706 I know that. But for any combination of the dice it will always have 4 cards among 52 that satisfies. Separate the dice odds is just unnecessarily complicate the problem.
I think there's the factor of the order of events. If you roll the dice first to get a number and then draw the card, then it's the 1/13. If you draw the card and _then_ roll the dice, now the odds are different and do depend on the dice probabilities.
I got something weird going on. I got r correct but i got R=sqrt13 because I used (y+r)²+2²=4² or (y+sqrt3)²+4=16 or y²=12-3 or y=sqrt9 This would give me y=3. When I correct it by saying y=r, the new formula gives 2(r²)+4=16 or 2(r²)=12 or r²=6 and since r=sqrt3, r²=3, making 3=6 which is impossible.
Google AI saw Andy’s side splitter theorem drawing and decided it looked important, so it put a box around it.
@@Crime_LabexCiting*
@@caseygreyson4178 :):):):)
Awesome 😂😂😂
How exciting!!
😂😂😂😂
I googled it but didn't get back your pic, or any AI result for that matter. That was in Firefox. So I decided to switch to Chrome (which I pretty much never use anymore) and indeed I got exactly the same result as your did. How exciting! 😃
Whats the difference between google and chrome
@@anghme28ang11Chrome is the browser, acting as your way to the internet. Google is just the search engine that you search web pages. Chrome uses Google as it’s default search engine, but you can change it to something like DuckDuckGo for more privacy, while using chrome as it’s browser.
Chrome is a browser made by Google company.@@anghme28ang11
@@anghme28ang11 'Google' is a corporation which makes a search engine called 'Google Search' (often shortened to 'Google') and 'Google Chrome' which is a web browser (often shortened to 'Chrome').
This is all academic because Firefox and Brave are both better than Chrome anyway.
@@anghme28ang11Crome is Google’s web browser, Google is the search engine run by Google (the company) - so when searching on Firefox (a different web browser), even though they used Google (the search engine), they didn’t get Google AI’s summary, as that seems to be attached to the Crome browser, not the Google website.
1:26 That's super cool! And I love how your talk went off on...... a tangent. ;-)
I've always loved the careful way you say little
It gives me the ik
HOW EXCITING 🗣🗣🔥🔥💯
I love how happy Andy was to see his image used by Google. How exciting.
Not that I’m distracted, but the three doors behind him, and one of them being open.. deeply disturbs me.
like a sudden jumpscare would just pop out.
maybe there's goats behind the other two doors.
@@paulvansommerenso the open door has a car behind it? That can’t be right.
@@tortinwall 67% chance! 😁
Googling the side splitter theorem didnt show AndyMath for me :(
Thank you for checking!
It did for me.
Google may bring up different images at random. Worked for me. Maybe give it another try.
@@richardl6751 It is because of location, as an Indian, I was shown from Byjus, Vedantu, etc cause there are a lot of websites like that.
Omg this is the first time I’ve actually been able to solve the problem before Andy explained it although I did do it a different way but it’s geometry so it doesn’t matter which way it’s solved
No need for "side splitter theorem". The triangle contained within the orange semicircle is an equilateral triangle in its own right, with side 2 and height root 3.
Then you can draw a line from the midpoint of the top triangle’s base to one of the end points of the bottom triangles base (the length of this = R), this line forms a triangle, you know that the shortest side = 1, and the third side = 2, the angle between these two sides is 120 degrees (180 - the 60 degrees from the top triangle), you can then use the cosine rule to work out R.
R = sqrt(1^2 + 2^2 - 2(1x2)(cos(120 degrees)))
R = sqrt(5 - 4cos(120))
R = sqrt(5 -(-2))
R = root 7
this video looks important, lets put a box around it. and thats our solution. how, exiting.
The question is easy because we directly assumed the midpoint of the smaller equilateral triangle as center, proving it is actually the center and infact the center for both semicircles seemed a lot more challenging.
It is perpendicular to the diameter(perpendicular as it is in an isosceles triangle).
Euilateral trianle means all sides are 4, so each hashed segment is 2. Split the 🔺️ in ½ and you get 2 30/60/90 Special 🔺️s. Once you appreciate that, the math takes care of itself
I got the same picture to your link, too. LOL So cool.
Thank you for providing the full puzzle in the thumbnail so I can give it a go before checking answers :D
I googled it and it linked to your site! Well done sir.
After watching this I first searched S.P. Theorem in the incognito mode, no result. Then in normal chrome and got a picture but different and after clicking on that image and scrolling below there was your image after few images. So, i think due to better answer of yours it is either showing on top or in top top 10 image results alongside.
Thanks Andy.
How exciting!!!🖤
You can easily get the orange radius by knowing that the hypotenuse is 2 and the and the base of the right triangle is half of that since it’s an equal lateral.
It shows your AndyMath picture for me!
i used Cosine Rule to solve for the larger circle. Good one on the Pythagoras theorem for the larger circle as well.
I googled side splitter theorem and the search result was the same as you got in the video!
I had a different solution to this problem. I proved similarity by angles and used pythagorean theorum on the smaller triangle. Using the height i found the area of the smaller semicircle. Then i used pythagorus to find the entire larger triangles total height and subtracted the radius of the smaller circle. Then i finally used the same method for the area of larger semicircle.
I tried it a different way and got messed up lol. I tried to take the bottom left corner of the triangle, form it into a right angle with the other end going to the bottom midpoint of the circle, and then a straight line up (the hypotenuse) going straight up and bisecting the circle. Top was then 30° so it's a 30/60/90 triangle. I messed it up because i somehow got 27π/4 as the area of the bottom semicircle lol.
The worst part is that i also tried to do a normal right angle thing like you did with the 2 and √3, but in my mind i thought it was going to be a 45/45/90 and when it wasn't i just abandoned that idea for some reason lol.
Great video :) the side splitter theorem thing shows up on google when you use chrome but not firefox it seems, odd how its browser dependent lol
notice how the triangle inside the orange semicircle must also be equilateral, and notice further that the smaller triangle in the orange semicircle can fit 4 times into the complete triangle.
then you can simply pythagoras your way to the solution.
I really should go to sleep
The side splitter theorem is also called Basic Proportionality Theorem
If you ever decide to start a podcast where you just talk about random stuff I'd listen to it. You have a nice voice.
hey! can you solve some hard-ish equations too? id like to see you solve the equations. it would be great.
The theorem worked for me, saw the image!
4:26 Wouldn't the probability of that happening just be 1/13? Because even though the dices have different probabilities for each number, the probability of the card matching it would always be the same, and there are 13 options
The minimum value you can get by rolling 2 die is 2 (2 ones), and the max is 12 (2 sixes)
So an ace (corresponding to 1) and a king (corresponding to 13) would never match.
Taking that into consideration, the 1/13 chance would probably not work
@@kshitijbenedict8431 I know but doesn’t matter, think like this: you roll the dices and it shows a number from 2 to 12, then from the 13 different card options, only 1 of them would match the dices. Hence the 1/13
@@kshitijbenedict8431 It is 1/13! whatever you roll doesn't matter, there is 36 dice combinations and each has a 1/36 chance. Even if you calculate it separately. Each one of those 36 combinations has the exact 1/13 of a match with the card. (i.e. (1/36 * 1/13) *36). You can do it the hard way of getting the probability of each possible number (2,3,4,5,6,7,8,9,10,11,12) but that would be pointless.
@@jalnaqi exaclty
@@miguelpinto5482thanks for the explanation! I'm a bit messy with probability, so I might use the table he made in the new vid
Totally came up for me!
How come you (almost) always root the squares when the formulas need the squares? Is there a reason to go through the extra steps?
I used little r later in the problem, that is why I rooted it. I didn't root Big R since I didn't need it later in the problem. I hope that makes sense.
Your diagram came up when I googled sidesplitter theorem.
Why use side splitter? Top triangle is equilateral so you know splitting it in half gives you 30-60-90. Sides are then 1, 2, and root 3. Seems quicker than side splitter theorem (but lacks AI reference to Andy Math).
at 1:02 why is the side 4? it is an isosceles triangle, not necessarily an equilateral triangle. Or did I miss something?
The image doesn't show it but the description states its an equilateral triangle.
@@AndyMath Yep, now I see it. My bad.
Thanks sir,,, i had the same question @@AndyMath
Ah yes , side splitter theorem....the thing we learnt just 4 months ago.
Only that it is for us called "Basic proportionality theorem" or BPT for short.
Not me at 12 at night watching this trying to concentrate 😅
Awesome though 😊
hey uh Andy, for your next video about probability, how do you roll a 13 if you got two dice? Is there something I'm missing? You did say King = 13
Since it's about probability, he'll just factor in the zero probability of matching a king, I guess.
Good shit
Your image didn't show up for me in a general search, but it was #4 in google images
Wow!!!
I'm confused why at 1:43 the side can be split from 4 into 2 and 2. What guarantees that the side split line between the half circles is splitting the side exactly in half?
The hash marks on the diagram denote line segments of equal length.
How do know the center coincide
I also got 5π, but using basic trigonometry!
Using Chrome on iPad, searched sidesplitter theorem and got many results and one of them was your site, but not AI result like yours.
Next video: isnt just 4/52? (1/13 or 7,69%)
A pair of dice do not have an equal probability to produce a number between 2 and 12. To roll a 2, you can only have 1-1, just as for a 12, 6-6. But for a three, you can have 1-2 or 2-1, for an eleven 5-6 or 6-5, and so on up to rolling a 7 (1-6, 2-5, 3-4, 4-3, 5-2, 6-1).
So, the odds of rolling a 7 and picking a 7 are (1/6)(1/13), or 1/78. But the odds of rolling a 2 and picking a 2 are (1/36)(1/13), or 1/468, the same as rolling a 12 and picking a queen. You can easily calculate the other potential outcomes simply by looking up the probability of rolling a number with 2 dice and multiplying it by the common (1/13) for then pulling the right card.
I suppose one trick could be that you have a 100% probability of getting a number between 2 and 12 with the dice, and therefore the 1/13 for then pulling a card that matches the specific number, but that feels like a cop-out.
@keith6706 I know that. But for any combination of the dice it will always have 4 cards among 52 that satisfies. Separate the dice odds is just unnecessarily complicate the problem.
@@pedroamaral7407 I know, but like I said that feels like cheating.
Exactly, because no matter what the dices say, there will always be a 1/13 chance that the card number is the same
I think there's the factor of the order of events. If you roll the dice first to get a number and then draw the card, then it's the 1/13. If you draw the card and _then_ roll the dice, now the odds are different and do depend on the dice probabilities.
Why does bisecting triangle still apply two sides splitter theorem? It's referring to its legs not too bisecting it.
1:40 how exciting!
Does anybody know the software he’s using for the slideshow? Thanks
Eazy pizy
Didn’t knew Thales theorem had other names
The video is 4:44 long! About an equilateral triangle! Of sides length 4!
Google in Germany says MathBitNotes for the Side Splitter Theorem
I got something weird going on. I got r correct but i got R=sqrt13 because I used
(y+r)²+2²=4² or (y+sqrt3)²+4=16 or y²=12-3 or y=sqrt9
This would give me y=3.
When I correct it by saying y=r, the new formula gives
2(r²)+4=16 or 2(r²)=12 or r²=6
and since r=sqrt3, r²=3, making 3=6 which is impossible.
Cool video! What games do you play?
My google AI Overview is showing your pic
How Exciting
not for me in safari, did the search for side splitter theorem bring the same results as yours :)
If I type Sidesplitter theorem I got your image, but if I google Side Splitter Theorem I do not get your image
That's an interesting test!
Andy meth
for me, it did not show up on the ai, but it was the 3rd image on google images
Math andy over here 😂
I did using law of cosines
I dont get it.
How do you know that the triagle sides are equal?
First sentence in the description states the triangle is equilateral.
Oh sh.t. I'm embarrassed. Thanks
So citrusy, we got an orange and a lemon in the problem, and we can make 5 pies out of them. 🍊🍋🥧🥧🥧🥧🥧
How the hell did i get -10.2
Solve this question:- 3^m-3^m =65 m=? Please
I think you need a few brackets in there, as the question makes no sense as stated.
Yup, I got the graphic when I googled!
OMG FIRST TIME I DID AN ANDY MATH PROBLEM ON MY OWN. IDK WHAT SIDE SPLITTER THEOREM WAS BUT I HAD MY OWN WAYS. EEEEEE
No Andy msth side splitter theorem diagram for me 🙄
I didn't get your pictures as the first pictures you see but if you click one of those images, Andymath is right there in the related pictures
Me
first
how exciting!
how exciting!
how exciting!
komenty na tt takich jak ja albo gorszych pustych ignota
It worked for me,respond or like if it also worked for u.(not obrigatory)
Probability question, answer is 1/13.
Probability of matching a King is 0.
For your next video, the answer is 50%. You're either going to pull a card out with the same value as the dice, or you're not. 🤷😆