@@aryapanjidwiputra554 3blue1brown did that one, which I'm assuming is what started this whole chain. ruclips.net/video/OkmNXy7er84/видео.html&ab_channel=3Blue1Brown
Like the new animations! There's a small mistake starting 2:24. The right side of the upper right equation should be 5^2 + (11 + x)^2, not 5^2 + (11 + x^2). However the calculations are done correctly, so it's just a display error.
@@ndreyey908 same man iam already confused soul and seeing that really made me question my life like do you know the distributive property of square and additions ?
Tried a bunch of new animations (and sounds) in this one. Feedback? Thanks so much for viewing! If you liked it, great! If you didn't like it leave a dislike and tell me why! :)
Just one small thing, and it is small. In the equation at the top right of the screen, you put (11+x^2) instead of (11+x)^2. Other than that it was fantastic, loved it
You actually explained this so well. Right now I’m an undergrad student doing math, but I’ve never really taken much geometry. So the fact that I could understand this should make you proud
1. I was going to do an A1 on area between curves using literally the same title. You beat me to it, so back to the drawing board. 2. The animations, sounds, conceptual layout, and geometric renderings were spot on. I know how time consuming these videos can be, - especially geometry - so your work is greatly appreciated!
I actually used an entirely different solution, based on the Euler line. First, let's construct line AM. M is the midpoint of BC, so AM is a median and therefore passes through the centroid. For any triangle the orthocenter, centroid and circumcenter lie on the same line, therefore the centroid will be the point of intersection of AM and HO. We will call this point X. The Euler line's property that 2 XO = XH means that we can determine XH to be 22/3, since HO is 11. We can now use congruence of AMF and AXH to find that AH = 10. This allows to find AO using Pythagoras through AH and HO, which turns out to be sqrt(221). AO is equal to BO because radius, therefore we can once again find BM using Pythagoras through BO and OM. BM = sqrt(221-25) = 14, therefore BC = 28.
We can also find AH=10 using the fact that the reflections of the orthocenter in the sides lie on the circumcircle - denoting reflection of H through BC by H' we have HF=FH'=5 so HH'=10. But since OH is perpendicular to the chord AH' H must be its midpoint and thus HA=10 as well.
As we know o and H we know it's euler line that means we can able to find vertexA also we know radii of Circumcircle as radi of nine point circle is half so we can find all three vertex by contructing bigger circle(circumcircle) ,
Great video...I always hate geometry due to all of its constructions, but how you did it, now I am starting to realize how to approach a problem, like spotting the right triangles etc...Keep on
Very nice video. I personally would had try to avoid the analytic techniques with slopes using the clasical triangle similarity (ABF and HCF), getting the same equation.
This was a great video, but, no offense, I don't think a video this short just showing the solution does justice to how hard the problem really is. Maybe in future show some lines of thought that look promising but ultimately dead-end so it feels more realistic to the "real experience". That's just my two cents, but great solution nonetheless!
There's a theorem that says that AH=2OM. We can easily calculate the radius of the circle via Pythagorean theorem for triangle HOA and calculate BM thereafter.
you can also do it totally geometric without any real equations: let H' be the reflection of H over BC, then by pythagoras we can find the radius since we know HH' and OH, and then we can look at triangle BMO to find length MB and done An alternative way would be to introduce N9, the midpoint of the 9-point circle, since you know that it lies on the midpoint of line OH, you can find the radius of the 9-point circle which is half the radius of the circumcircle and finish the same as before
Here I provide a faster solution to solve this problem (provided that you know the necessary results): Since HFMO is a rectangle, the Euler line OH is parallel to BC. The centroid G lies on OH and AM, so AH/HF=AG/GM=2. This implies AH=10. Applying Pythag twice, AH^2+OH^2=AO^2=BO^2=BM^2+MO^2, which gives BM=14 so BC=28.
This was really cool but im kinda confused on the step where he multiplied the slopes, does anytime you multiply 2 altitudes of a triangle always equal -1?
No, they equal -1 because the two slopes were perpendicular. This happens anytime you take the product of perpendicular slopes. (Take 3 and -1/3 for example)
Note that the orthocenter H and the circumcenter O are collinear with the centroid, and the centroid is always 1/3 the way from a base to the opposite vertex. So if you know that, you get y=10 for free.
Do exploring this problem, I came across the 9-point circle. The center of the 9-point circle is N, which is midway along HO. And it has some interesting properties. The points F and M are both on the 9-point circle, as is a point halfway along the line segment AH. Let's call that point P, so because of how circles work FH is the same length as HP, and because P is halfway along AH, AH is twice the length of FH. Further because of intersecting chords we can say AH^2=(r-11)(r+11) where r is the radius of the circumcircle, and 11 is the length OH. r=sqrt(221). BM^2=r^2-OM^2=221-25=196. BM=CM and BM+CM=BC=28
Umm another solution can be reflecting H in BC and then using the fact that it lies on the circumcircle to directly calculate radius^2 = OH^2 + 10^2 = 11^2 + 10^2 = OM^2 + (BM)^2 = 5^2 + (0.5 BC)^2 which gives BC = 28 It obviously assumes that one knows the reflection property of orthocentre , but other than that , it works
Wouldn't you also be able to solve this using calculus? Since O is the radius, and you just figure out the rate of change of the diameter as O moves down 5 units?
Alternate (arguably easier) solution: Extend AF to meet the circle at G. Let FG = y and BF = x Using the fact that H & M are the midpoints of AG and BC (since centre always bisects the chord), and the fact that HF = 5 & MF = 11, We can see that AF = y + 10 and FC = x + 22 Chords AG and BC intersect. Using law of intersection of chords we get x*(x + 22) = y*(y + 10) Observe that angles BAF and BCH are equal as both are part of right triangles sharing a common angle (angle ABC). Also, angles BAF and FCG are equal (angles in same segment) Using these two facts we get angles HCF & FCG are equal. Therefore using RHS postulate we get that triangles FCH and FCG are congruent. Therefore FG = FH = 5 In other words y = 5 So using the earlier equation we get x*(x + 22) = 5*(5 + 10) x*(x + 22) = 75 It can be seen that x = 3 Thus BC = x + (x + 22) = 28
Yeah, sounds are a bit too loud, and I think it would be better to have it so when one equation turns into another, the new one starts directly below it, then goes up top at the start of the next calculation, rather than appearing at the top immediately. Kinda take advantage of the whole way it's done on paper or a board, where the logic goes down until you run out of room, then on the next "page" the facts discovered appear at the top
@@somaannn i did take 16 , from the 5 and the 11. divide it by a factor of 2. 8 now has the total of 5 + extra space (x = 3) (11 x 2) + (3 x 2) = 28. its soo simple
To think this is the easiest problem....It's easy enough to understand the solution with an education in geometry but to actually find those relationships takes a lot of creative and insightful thinking.
My 1st step was to look at triangles ABF and CBH. They have angle B in common as well as both having a 90 degree angle (angles H and F) so angle A must equal angle C and the triangles are similar triangles. Next I looked at triangles CBH and CHF. They have angle C in common and both have a 90 degree angle so angle B must equal angle H which means they are similar triangles as well. it also means triangles ABF and CHF are also similar triangles. And that means that tan A=tan C and so BF/AF=HF/CF and BF*CF=AF*HF. With HF=5 and CF=22+BF that gives BF(22+BF)=5AF. Then I drew lines AO and OC to get AO²=AH²+11²=5²+CM². With CM=11+BF that gives AH²+11²=5²+(11+BF)²=5²+11²+BF(22+BF) which simplifies to AH²-5²=BF(22+BF). With two values for BF(22+BF) I then got 5AF=AH²-5² and with AF=AH+5 that becomes 5AH+5²=AH²-5² and so AH²-5AH-50=0 which means AH=10. So BF(22+BF)=10²-5²=75 and so BF²+22BF-75=0 which gives BF=3 and BC=2(BF+11)=28.
why are you using slopes instead of similar triangles?! (CFH and AFB) CF/HF=AF/FB => (22+X)/5=(5+y)/X and you get the same result by using only pure geometry
Alternative solution - let N be the midpoint of OH. It's a well known fact that N is the nine point centre of ABC and the radius of the nine point circle is half the radius of the circumcircle. F and M both lie on the NPC, and by Pythagoras, NF=sqrt((11/2)^2+5^2)=sqrt(221)/2. Hence OB=OC=sqrt(221). So MC=sqrt(221-5^2)=sqrt(196)=14. Hence BC=2MC=28.
🙏🙏Sir, Even though I practice 4 to 5 hours a day but I'm still weak, I'm too slow in doing algebra problems. I love maths and I want to be good in it please help. 🙏🙏🙏🙏
Keep going! If you continue to practice like this, mastery will come before you know it. Continue to use RUclips and other resources to help. You can do this!
#USA another book for the student world wide again 5th time no denying this boy now said study's #UsGoverment #Worldhistory that bitch ex is that manw yash let me check him out
@@BriTheMathGuy I was taking about that we need to form the equations carefully.That's why I said that we need to be attentive to do these type of problems
3b1b-Hardest problem from hardest test Bri-Easiest problem from hardest test Myd- Problem from the hardest test Edit:bprp and myd both made easy problem from hardest test
Here's my approach: It is well known that the center of the Feuerbach circle is at the midpoint of OH, (let's call it F) and its radius is R/2. M is the midpoint of BC so it's on the Feuerbach circle, which means FM=R/2. By the Pythagorean Theorem (in triangle FOM) FM=sqrt(221)/2, so R=sqrt(221), this means, that OC=sqrt(221) Now using the Pythagorean Theorem once again in triangle OCM we get CM=14, so BC=28.
I'm 15, I did it too! XD this was so easy... Although instead of all that slope and stuff I realised that 🔼ABF and 🔼CHF are similar so AF/CF = BF/HF I labelled x and y just as you did, so when I saw your labelling after I solved, I was like "You cheated"! Hahahaahh!!!
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Now we just need “the easiest question on the easiest test” and “the hardest question on the easiest test” and we will complete the dynasty!!
Gotta collect em' all!
We still need the hardest question of the hardest test (cmiiw)
@@aryapanjidwiputra554 3blue1brown did that one, which I'm assuming is what started this whole chain. ruclips.net/video/OkmNXy7er84/видео.html&ab_channel=3Blue1Brown
What is the easiest test?
The easiest question on the easiest test is filling in your name.
Like the new animations!
There's a small mistake starting 2:24. The right side of the upper right equation should be 5^2 + (11 + x)^2, not 5^2 + (11 + x^2). However the calculations are done correctly, so it's just a display error.
🤦
the difference between you and me is that you understand this and i don't
I first taught that I was having hallucinations lmao...
@@ndreyey908 same man
iam already confused soul and seeing that really made me question my life
like do you know the distributive property of square and additions ?
Tried a bunch of new animations (and sounds) in this one. Feedback?
Thanks so much for viewing! If you liked it, great! If you didn't like it leave a dislike and tell me why! :)
Love the new animations!
Just one small thing, and it is small. In the equation at the top right of the screen, you put (11+x^2) instead of (11+x)^2. Other than that it was fantastic, loved it
I still miss you writing backwards coz you don't make those strange expressions anymore lol
It's amazing
They're great
You actually explained this so well. Right now I’m an undergrad student doing math, but I’ve never really taken much geometry. So the fact that I could understand this should make you proud
1. I was going to do an A1 on area between curves using literally the same title. You beat me to it, so back to the drawing board.
2. The animations, sounds, conceptual layout, and geometric renderings were spot on. I know how time consuming these videos can be, - especially geometry - so your work is greatly appreciated!
I actually used an entirely different solution, based on the Euler line. First, let's construct line AM. M is the midpoint of BC, so AM is a median and therefore passes through the centroid. For any triangle the orthocenter, centroid and circumcenter lie on the same line, therefore the centroid will be the point of intersection of AM and HO. We will call this point X. The Euler line's property that 2 XO = XH means that we can determine XH to be 22/3, since HO is 11. We can now use congruence of AMF and AXH to find that AH = 10. This allows to find AO using Pythagoras through AH and HO, which turns out to be sqrt(221). AO is equal to BO because radius, therefore we can once again find BM using Pythagoras through BO and OM. BM = sqrt(221-25) = 14, therefore BC = 28.
We can also find AH=10 using the fact that the reflections of the orthocenter in the sides lie on the circumcircle - denoting reflection of H through BC by H' we have HF=FH'=5 so HH'=10. But since OH is perpendicular to the chord AH' H must be its midpoint and thus HA=10 as well.
Idk bro I used a ruler.
why i felt the need to watch this in year 9 i have no idea
I quite like the animations, the sounds are a little distracting.
As per usual, great explanations. Loving this channel.
Noted! Thanks very much for the feedback and have a great day!
I need to solve this on my own 😎 thank you for the motivation 💙
You got this!
As we know o and H we know it's euler line that means we can able to find vertexA also we know radii of Circumcircle as radi of nine point circle is half so we can find all three vertex by contructing bigger circle(circumcircle) ,
Great video...I always hate geometry due to all of its constructions, but how you did it, now I am starting to realize how to approach a problem, like spotting the right triangles etc...Keep on
Thanks so much! Have a wonderful day!
Please do a video on must know series. Thank you 💫
The animations are really wonderful. 😁 I wonder what did he use for them.
Glad you like them! (I edit in Final Cut Pro)
Finally a question whose every step of the solution I could understand
Thanks for watching! :)
Very nice video. I personally would had try to avoid the analytic techniques with slopes using the clasical triangle similarity (ABF and HCF), getting the same equation.
Maybe I should have done it that way! Thanks for watching/commenting!
Whenever im solving geometrical problems, i never come across the idea to use slopes to find an unknown value, so this is new to me
Channel getting better everyday
Thanks so much! Have a great day!
Nice solution - doesn´t strike me as so easy, but I´ve never seen that exam
nice! wanna see more geometry on your channel
Thanks! I'll do my best!
This was a great video, but, no offense, I don't think a video this short just showing the solution does justice to how hard the problem really is. Maybe in future show some lines of thought that look promising but ultimately dead-end so it feels more realistic to the "real experience". That's just my two cents, but great solution nonetheless!
I think that's some great feedback! Thanks very much and I'll do my best in the future!
There's a theorem that says that AH=2OM. We can easily calculate the radius of the circle via Pythagorean theorem for triangle HOA and calculate BM thereafter.
Thinking in an “Euclidean” way, I would rather demonstrate that ΔCHF ~ ΔABF, which means CF/FH = AF/FB, namely
(x+22)/5 = (y+5)/x
you can also do it totally geometric without any real equations:
let H' be the reflection of H over BC, then by pythagoras we can find the radius since we know HH' and OH, and then we can look at triangle BMO to find length MB and done
An alternative way would be to introduce N9, the midpoint of the 9-point circle, since you know that it lies on the midpoint of line OH, you can find the radius of the 9-point circle which is half the radius of the circumcircle and finish the same as before
I see you changed geometry problem to just problem to clickbait the geometry haters xD
They'll never see it coming 😏
Here I provide a faster solution to solve this problem (provided that you know the necessary results):
Since HFMO is a rectangle, the Euler line OH is parallel to BC. The centroid G lies on OH and AM, so AH/HF=AG/GM=2. This implies AH=10.
Applying Pythag twice, AH^2+OH^2=AO^2=BO^2=BM^2+MO^2, which gives BM=14 so BC=28.
This was really cool but im kinda confused on the step where he multiplied the slopes, does anytime you multiply 2 altitudes of a triangle always equal -1?
No, they equal -1 because the two slopes were perpendicular. This happens anytime you take the product of perpendicular slopes. (Take 3 and -1/3 for example)
Note that the orthocenter H and the circumcenter O are collinear with the centroid, and the centroid is always 1/3 the way from a base to the opposite vertex. So if you know that, you get y=10 for free.
Good video and a nice problem
Glad you liked it!
Great problem there, and considering I don't even remember the circumcircle rules too
Thanks 👍
No problem 👍
Can you provide some stuff for jee.🙏
Do exploring this problem, I came across the 9-point circle. The center of the 9-point circle is N, which is midway along HO. And it has some interesting properties.
The points F and M are both on the 9-point circle, as is a point halfway along the line segment AH. Let's call that point P, so because of how circles work FH is the same length as HP, and because P is halfway along AH, AH is twice the length of FH. Further because of intersecting chords we can say AH^2=(r-11)(r+11) where r is the radius of the circumcircle, and 11 is the length OH. r=sqrt(221).
BM^2=r^2-OM^2=221-25=196.
BM=CM and BM+CM=BC=28
Umm another solution can be reflecting H in BC and then using the fact that it lies on the circumcircle to directly calculate radius^2 = OH^2 + 10^2 = 11^2 + 10^2 = OM^2 + (BM)^2 = 5^2 + (0.5 BC)^2 which gives BC = 28
It obviously assumes that one knows the reflection property of orthocentre , but other than that , it works
Hi sir . Can u explain NOMR , LP metric
Thanks
Why is the product of m1 and m2 -1.
You said they were perpendicular but that’s to the triangle not to the lines themselves. I’m confused.
Wouldn't you also be able to solve this using calculus? Since O is the radius, and you just figure out the rate of change of the diameter as O moves down 5 units?
not really since that changes everything else aswell kinda.
Your channel was already neat before, now you're making it incredible. Keep it up!
Btw, what program do you use to make those animations?
Thanks so much! I use Final Cut Pro.
Can anyone explain the part where he said the product of m1 and m2 should be -1? I'm confused but i will really appreciate an answer
If there are two lines perpendicular to each other the product of their gradient is -1 (gradient is shown by m )
Alternate (arguably easier) solution:
Extend AF to meet the circle at G.
Let FG = y and BF = x
Using the fact that H & M are the midpoints of AG and BC (since centre always bisects the chord), and the fact that HF = 5 & MF = 11, We can see that AF = y + 10 and FC = x + 22
Chords AG and BC intersect. Using law of intersection of chords we get x*(x + 22) = y*(y + 10)
Observe that angles BAF and BCH
are equal as both are part of right triangles sharing a common angle (angle ABC).
Also, angles BAF and FCG are equal (angles in same segment)
Using these two facts we get angles HCF & FCG are equal.
Therefore using RHS postulate we get that triangles FCH and FCG are congruent.
Therefore FG = FH = 5
In other words y = 5
So using the earlier equation we get x*(x + 22) = 5*(5 + 10)
x*(x + 22) = 75
It can be seen that x = 3
Thus BC = x + (x + 22) = 28
Yeah, sounds are a bit too loud, and I think it would be better to have it so when one equation turns into another, the new one starts directly below it, then goes up top at the start of the next calculation, rather than appearing at the top immediately.
Kinda take advantage of the whole way it's done on paper or a board, where the logic goes down until you run out of room, then on the next "page" the facts discovered appear at the top
Well said! I'll try to do my best with this in the future!
If we can find the exact value of 0/0.then,we can also find the momentum of ( C )
I don’t understand how this shit is gonna help me in my future career
How do we know x is the same as GC?
Wait how do you know that x on the left is the same as x on the right
Wow, I'm honestly impressed with myself. I was able to solve it!
And yes I'm going to flex on you because I'm in 9th grade.
You are so awesome!
You are!! 😁
i agree
@@particleonazock2246 hello
@@aashsyed1277 hello again
@@particleonazock2246 oh so you also comment on this channel =D
I've lost you at "Here is the problem.."
Same question was in my jee practice sheet
You can do that way easier though. Just measure the distance between O and the orange triangle side above it.
Answers by accurate drawing are not accepted. You need to calculate it.
@@somaannn i did
take 16 , from the 5 and the 11. divide it by a factor of 2. 8 now has the total of 5 + extra space (x = 3) (11 x 2) + (3 x 2) = 28. its soo simple
This question is so easy when you think logically ;-;
To think this is the easiest problem....It's easy enough to understand the solution with an education in geometry but to actually find those relationships takes a lot of creative and insightful thinking.
My 1st step was to look at triangles ABF and CBH. They have angle B in common as well as both having a 90 degree angle (angles H and F) so angle A must equal angle C and the triangles are similar triangles.
Next I looked at triangles CBH and CHF. They have angle C in common and both have a 90 degree angle so angle B must equal angle H which means they are similar triangles as well. it also means triangles ABF and CHF are also similar triangles. And that means that tan A=tan C and so BF/AF=HF/CF and BF*CF=AF*HF. With HF=5 and CF=22+BF that gives BF(22+BF)=5AF.
Then I drew lines AO and OC to get AO²=AH²+11²=5²+CM². With CM=11+BF that gives AH²+11²=5²+(11+BF)²=5²+11²+BF(22+BF) which simplifies to AH²-5²=BF(22+BF).
With two values for BF(22+BF) I then got 5AF=AH²-5² and with AF=AH+5 that becomes 5AH+5²=AH²-5² and so AH²-5AH-50=0 which means AH=10. So BF(22+BF)=10²-5²=75 and so BF²+22BF-75=0 which gives BF=3 and BC=2(BF+11)=28.
why are you using slopes instead of similar triangles?! (CFH and AFB) CF/HF=AF/FB => (22+X)/5=(5+y)/X and you get the same result by using only pure geometry
Oh! Nice!
Alternative solution - let N be the midpoint of OH. It's a well known fact that N is the nine point centre of ABC and the radius of the nine point circle is half the radius of the circumcircle. F and M both lie on the NPC, and by Pythagoras, NF=sqrt((11/2)^2+5^2)=sqrt(221)/2. Hence OB=OC=sqrt(221). So MC=sqrt(221-5^2)=sqrt(196)=14. Hence BC=2MC=28.
Why is there no altitude coming from vertex B?
This was just the way the problem happened to be proposed. Have a great day!
you killed my brain
🤯
I got the answer by estimating the length of BF by looking at the size difference from FM and BF and made a estimated guess and got 28. Lucky
Make more videos like these, geared toward a beginner audience..
I'll do my best! Thanks for watching!
You forgot to mention during the problem statement that F is the foot of the altitude from A.
Great video anyways!
🙏🙏Sir, Even though I practice 4 to 5 hours a day but I'm still weak, I'm too slow in doing algebra problems.
I love maths and I want to be good in it please help.
🙏🙏🙏🙏
Keep going! If you continue to practice like this, mastery will come before you know it. Continue to use RUclips and other resources to help. You can do this!
@@BriTheMathGuy Thankyou Sir I'm very grateful.
Amazing video but there's a mistake 2:24 on top right corner. But we can ignore them because your content is really good.
What is the mistake
@@arinroday302 it will be 5^2 + (11+x)^2 not 5^2 + (11+x^2) but calculations further are correct so it doesn't matter.
@@mathevengers1131 hello
@@aashsyed1277 hello
@@mathevengers1131 never expected you will be here
easier way to find y wouldve been: the point where altitudes meet in a triangle divides them into a ratio of 2:1, meaning y=2*5
How do you know the relation is 2/1
or just use a ruler
alternatively y = 10 by euler line
when i saw the name brithemathguy i read it as "bribethemathguy"
Maybe it cuold be solved with eulers line
🤔
May be you can use euler line
Maybe!
this is pretty normal with vietnamese highschool students
@@BriTheMathGuy oh i looking foward that you will have a video to solve IMO 2021
Now do the hardest problem on the easiest test
3 blue 1 brown made a video named "the hardest problem on the hardest test"
Yup I watch that too , but you are here too :)
I think you should gonna start a channel (if you have time)
@@pardeepgarg2640 i will i think at the end of 2021
@@aashsyed1277 :O be sure to tell me :)
@@pardeepgarg2640 ok :DDD
^^ An amazing video ^^
Love from india .....❤❤❤❤
Thanks for stopping by! :)
@@BriTheMathGuy do u know about jee advance...iit jee
Can't u just say that y and HF are related like 2/3 and 1/3?
Beans⁶
or you use a ruler
32.33591 something like that
#USA another book for the student world wide again 5th time no denying this boy now said study's #UsGoverment #Worldhistory that bitch ex is that manw yash let me check him out
woah
i was supposed to be smart...😂
Me in 8th Grade, Hmmm yes. I understand this...
Easy but need to be attentive...
I appreciate feedback! Could you tell me a little more about what you mean please?
@@BriTheMathGuy I was taking about that we need to form the equations carefully.That's why I said that we need to be attentive to do these type of problems
@@supramitra got it thank you!
The 1st is a bit tilted
Thank you for watching and commenting!
Psalms 34:8
Wait a min exams name is....????
Putnam!
I like watching your videos but I'm very bad in English
You sound pretty good to me! Thanks for the support!
3b1b-Hardest problem from hardest test
Bri-Easiest problem from hardest test
Myd- Problem from the hardest test
Edit:bprp and myd both made easy problem from hardest test
I like the video. But I think it would have been better if you paused and asked the audience to try and solve it themselves first.
Thanks for the tip! I'll do my best!
First to learn and comment...
Thanks for watching! Have a great day!
Here's my approach:
It is well known that the center of the Feuerbach circle is at the midpoint of OH, (let's call it F) and its radius is R/2.
M is the midpoint of BC so it's on the Feuerbach circle, which means FM=R/2.
By the Pythagorean Theorem (in triangle FOM) FM=sqrt(221)/2, so R=sqrt(221), this means, that OC=sqrt(221)
Now using the Pythagorean Theorem once again in triangle OCM we get CM=14, so BC=28.
WAT
Please put the Arabic translation
I'm sorry I don't have that capability!
What
i dont understand english, also dont understand maths, why im watching this?
You are handsome
Great to hear!
There is a more easier solution to it.
I'm 15, I did it too! XD this was so easy... Although instead of all that slope and stuff I realised that 🔼ABF and 🔼CHF are similar so AF/CF = BF/HF
I labelled x and y just as you did, so when I saw your labelling after I solved, I was like "You cheated"! Hahahaahh!!!
Hello baby
What a waste of energy and time
√100=10 7sin7=0•853085403 10^4=10000 π-π=0 8log in8-8login8=0