A satisfying geometry question - circle exterior to a triangle side

Рет қаралды 240,585

Күн бұрын

Thanks to Amit from India for the suggestion! A circle is tangent on the exterior of a side of a triangle and tangent to the lines through the other two sides. The tangent side length 7, and the other side lengths are 8 and 9. What is the radius of the circle?
0:00 Problem
1:22 Geometry
5:35 Trigonometry
Twitter thread (additional solutions)
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Пікірлер: 390

@zzstoner 2 жыл бұрын

Dark Mode: Activated. Nice. :)

@fishcanroll0 2 жыл бұрын

now its just organic chemistry channel

@ArunaVishwajith123 2 жыл бұрын

I am in Sri Lanka .... this is one of my favorite channels .... your service is very valuable .... I learned a lot from this channel ...Thank you PRESH TALWALKAR 😍😍

@Spartans-community 2 жыл бұрын

Gammac mageth favourite channel ekak dana hama ganama hadnw 🖤❤️

@user-ci5kh7pl8y 2 жыл бұрын

Hi. I'm from your neighbouring country.😁😁

@Spartans-community 2 жыл бұрын

@@user-ci5kh7pl8y India 🇮🇳 ✋

@IS-py3dk 2 жыл бұрын

@@user-ci5kh7pl8y Me too neighbouring country 😃 India

@xxdxx3776 2 жыл бұрын

@DoTheMath lmaooo

@pwmiles56 2 жыл бұрын

To avoid the cosine rule: make angle NOD=theta, angle DOM=phi. Then angle NOM = theta+phi By right triangle properties 3 = r tan(theta/2), tan(theta/2) = 3/r 4 = r tan(phi/2), tan(phi/2) = 4/r 12 = r tan((theta+phi)/2) Use tan sum formula 12 = r (3/r + 4/r)/(1 - 12/r^2) 12 = 7/(1-12/r^2) 12 - 144/r^2 = 7 144/r^2 = 5 r^2 = 144/5 r = 12/sqrt(5)

@Tiqerboy 2 жыл бұрын

That looks far more complicated than avoiding the law of cosines.

@pwmiles56 2 жыл бұрын

@@Tiqerboy Oh, I thought it brought out the symmetries in the problem. Thnx for comment

@dickson3725 2 жыл бұрын

No, Just use similiarity to avoid complexity

@philosophical_aspect. 2 жыл бұрын

@@pwmiles56 man i have an ezz solution Just imagine like its a half rectangle Formula for diagonal of a rectangle = √l²+b²find that then half it and you get your answer !!!!

@pwmiles56 2 жыл бұрын

@@philosophical_aspect. What are l and b?

@spdigital7752 2 жыл бұрын

I'm from India, 17 years old and I solved this question by my own made Formula within 5 seconds.

@nirmalaverma8127 2 жыл бұрын

Its taught in india in class 10/11 th. So it was easy for me. Its basically ∆/s-a

@mevadavraj4178 2 жыл бұрын

Stop boasting please it's very cringey, ofcourse question is easy but u don't have to tell this is taught in india std 10 or 11

@dickson3725 2 жыл бұрын

But obviously you didn't know how to derive it hahaha

@pravegrajput 2 жыл бұрын

@@mevadavraj4178 yes bro you are right , it is not taught in class 11 in India but in class 10 I am learning it right now ,

@pravegrajput 2 жыл бұрын

If you want just check class 10 cbse circle previous year questions , you will find it.

@mevadavraj4178 2 жыл бұрын

@@pravegrajput i am not saying it is not taught in std 10 i am saying stop boasting urself and also i am in std 10

@SAM00780 2 жыл бұрын

I got exact same question is my 10th prelims but only to derive until we find the length of tangent segments.. btw, u taught this much easier than my own teacher..

@Bugrick92 2 жыл бұрын

In Darkmode now, nice! Interessting puzzle :)

@SQRTime 2 жыл бұрын

Hi Bugrick. If you are interested in math competitions, please consider kzfaq.info/get/bejne/gd1ndsZ9yKm5lWg.html and other videos in the Olympiad playlist. Hope it will be useful 👍

@TheDigiWorld Жыл бұрын

There is another way to go at the problem: We start with the Heron's formula to calculate the area, that (as explained in the video) is √720 As the question stated that one side of the triangle is tangent to the circle while the other two sides also extend as tangents to the circle, such a circle is called an Escribed circle and there is a formula for the radius of such circle r = Area of triangle/(s - a), where a is the length of the side of the triangle that touches the circle (in this case, BC). Simply putting in the value we get r = √720/(12-7) which simplifies to 5.366 or 5.37

@falldamage2008 Жыл бұрын

Hello, there is a simpler way to do this. Area of triangle ABC = r1(s-a) r1 is the radius we need to find, s is the semiperimeter and a is length of side BC. By herons formula we can find the area of triangle and then equate it to the RHS

@pat7594 2 жыл бұрын

you may also know that the radius of the escribed circle is equal to : 2S/(8 + 9 - 7), where S is the area of the triangle (found with Heron's formula)

@gaurav7582 2 жыл бұрын

My maths olympiad exam is on 6th march love from india

@nutmegninja23 2 жыл бұрын

Who knew thanos was Indian

@qwertyuiop-pc2xw 2 жыл бұрын

Thanoswaram iyer

@notananimenerd1333 2 жыл бұрын

Ioqm

@Anonymous-kw7ls 2 жыл бұрын

@@qwertyuiop-pc2xw Thanoswamy muttuswammy iyyer.

@akhandanand_tripathi 2 жыл бұрын

@@Anonymous-kw7ls thanoswamy mutthuswamy venugopal iyyer

@kashishgautam7040 2 жыл бұрын

R= area of traingle/ semiperimeter - side touched by Circle

@Vishalkr12 2 жыл бұрын

Your way of explanation is great. Love from INDIA. 01-03-2022..1:58 PM

@shashwatvats7786 2 жыл бұрын

Nice question . Please give these types of question countinously. It refrains our mind. It took me more than five minutes to solve.

@spencer1980 2 жыл бұрын

The geometry portion was elegant. Good work.

@dee5559 2 жыл бұрын

Thank you, can't believe I went my whole life never having heard of that so-easy Heron formula oO

@justnowi8967 2 жыл бұрын

Loved the new dark theme ❤️

@user-ly5bc4xd2s 2 жыл бұрын

تمرين جميل جيد . شرح جيد واضح مرتب . شكرا جزيلا لكم استاذنا الفاضل والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .

@helpmefortheloveofshrek6623 2 жыл бұрын

I actually solved this! I’m so happy :) I ended up creating a square, made up of two right triangles. Using the fact that a square is made up of 4 right angles, I split up the square into two right triangles. I then did trig to solve for the bottom of the triangle, using sin(45)=x/7 and got around 5 ( I just rounded up a lot to make my life easier). Since it is a square, all of its sides must be the same, therefore the side that I made have the same length as the radius is also 5, making the radius 5. I suck at explaining, so I hope any of it make sense

@devphalswal419 2 жыл бұрын

Sorry to tell you but there is no way you can create a square in this fig and all sides of square are equal which is not possible, you might have ended up making a kite

@helpmefortheloveofshrek6623 2 жыл бұрын

@@devphalswal419 maybe. I’m not an expert in math, and I just did this for fun, so I don’t what can and can’t be done. I just did something that somehow worked, so make of that what you will🤷‍♀️

@helpmefortheloveofshrek6623 2 жыл бұрын

@@devphalswal419 Also, I would like to know your reasoning. Maybe I can learn something.

@animezoneamv9116 Жыл бұрын

@@helpmefortheloveofshrek6623 But please tell me what square you formed

@jeyakumarleveenth6528 2 жыл бұрын

Yesterday we got a similar question in Sri Lankan mathematics competition.But there triangle ABC was a right angle triangle.I solved the question,as I knew the formula [A=(s-a)r] where A-Area of triangle,s-triangle perimeter/2,a-length of side opposite to

@vashon100 2 жыл бұрын

2:44 From the 7-8-9 triangle (ie ABC), solving angle C and its complement are 58.41 deg and 121.59 deg. Then segment ND can be obtained from law of cosines (triangle NCD). Last step, triangle NOD with sides of r solved with law of cosines again.

@MattColler Жыл бұрын

Nice approach - without a calculator, cos(angle ACB) = 11/21, so cosine of the supplementary angle NCB is -11/21. This gives segment ND² = 192/7, and with angle NOD being supplementary again, you can get the correct radius of 12/√5.

@rahulpaul5539 2 жыл бұрын

I so much like your trigonometry method .......it's so easier than geometry

@kimjennie7937 2 жыл бұрын

Nice question👌🏼

@alaaalkhafaji2884 2 жыл бұрын

Excellent for you to this 👏question

@HungNguyen-ey9wm 2 жыл бұрын

Love these solutions ❤

@JohnJones-pu4gi 2 жыл бұрын

Just to add the missing bracket: r_n = TriangleArea / (s - side_n) is the radius of the excircle on side n this is easy to see by looking at triangles OAB, OAC, and OBC and r_0, the incircle radius is TriangleArea/s which is even easier to see using the same technique. Of interest then by multiplying these together Pi(r_i) [i=0..3] = TriangleArea^2

@neysantos2147 2 жыл бұрын

The two ways are amazing. Thank you!

@WoodyC-fv9hz 2 ай бұрын

~5.3665. I worked out 3 angles of the triangle given and side-shifted its side (labeled 7 units) so far to the left, until tangential with the circle, thus inscribing the circle within a trapezoid of 2 parallel sides. "Similar triangles" finds all inner angles of the trapezoid. Connecting the incentre with point C and D, finds another triangle (to the left of triangle ABC), sharing it's base with line BC. Two of its angles are bisectors of the complementary angles (with respect to 180°) of BCA and ABC. Looking at the new triangle, having its apex at the incentre, 2 of its angles are already known and also known is the length of the base (7). I constructed a line from the incentre (O) perpendicular to BC (point D), which resembles the altitude of the new triangle, being equivalent to radius r.

@paulgets2737 2 жыл бұрын

Welcome back, Presh.

@pilu1966 Жыл бұрын

thanks for boosting my love for maths!

@powersulca3033 2 жыл бұрын

I solved the exercise in the second way, it had not occurred to me to use areas, the first way was very good, greetings from Peru

@kennethchristianguiwo9407 2 жыл бұрын

You can use Area of ∆ = r(s-a) where s is the semi perimeter and a is the tangent line to the circle and is also a side of the triangle. This is the shortcut.

@luismuller6505 Жыл бұрын

And you then use heron's formula to get s = (7 + 8 + 9) / 2 = 12 r = A / (s-a) = sqrt(s * (s-b) * (s-c) / (s-a)) = sqrt(12 * 3 * 4 / 5) = 12 / sqrt(5). Multiply both the numerator and denominator by sqrt(5) to get r = 12*sqrt(5)/5 which is the correct result.

@xz1891 2 жыл бұрын

If u know so-called excircles 旁切圆, the 3 rads are, 2A/(a+b-c), 2A/(a-b+c), 2A/(-a+b+c), respectively

@newzero1000 2 жыл бұрын

escribed circles : r1 = A/( s-a) where A = rt { s(s-a)(s-b)(s-c)} and s = (a+b+c)/2

@harinandnk123 2 жыл бұрын

This is the creative logic

@honsthebronze 2 жыл бұрын

Yes exactly that's what I am saying 😂 We also had a problem like this in our geometry book in Iran He made suck a mess didn't he😂

@honsthebronze 2 жыл бұрын

@@newzero1000 exactly That's right

@242math 2 жыл бұрын

great job, thanks for sharing

@TheMemesofDestruction 2 жыл бұрын

Thank you!

@davspa6 2 жыл бұрын

AM = AN, because the circle will touch the opposite lines at the same distance from A. BM = BD, and CN = CD, same reason. Also BD + CD = 7, so BM + CN = 7. So then AM + AN = 9 + 8 + 7 = 24. So AM = AN = 12. Use the law of cosines to find that cosine of angle A is 96/144 = 2/3. So then tan of half that angle = r/12. r = 12 tan(arccos (2/3)/2) ≈ 5.367

@hinsontarrayo5822 2 жыл бұрын

in short to get the radius of the circle just use the radius formula of the TRIANGLE ESCRIBED TO CIRCLE: *first find the "S" by using Semi-Perimeter of the Heron's formula, S=(a+b+c)/2 *let "a" the side of the triangle near to the circle or tangent to the circle so our a is 7, b is 8 and c is 9 *S=(7+8+9)/2 S=12 *now use the radius formula of the TRIANGLE ESCRIBED TO CIRCLE, r=A/(S-a) "A" is the Area of the triangle and also equal to the Heron's formula, √S(S-a)(S-b)(S-c) *r=(√S(S-a)(S-b)(S-c))/(S-a) r=(√12(12-7)(12-8)(12-9))/(12-7) r=5.36656

@natashok4346 2 жыл бұрын

Thank you! The black screen is best for my eyes.

@eroraf8637 2 жыл бұрын

Sweet, merciful dark theme.

@gaelik10 2 жыл бұрын

Solving it through a coordinate plane is a little bit more tedious but it can be done. We need 6 functions and 6 variables, at least in the way I did it. These are the equations: y=(1÷1.626)(x-z)+h y=-1.626 (x-9) h=sqrt((z-x)^2+(h-y)^2) o=1.118a o=(-1/1.1188)(a-z)+h h=sqrt((a-z)^2+(h-o)^2) Where: ## h and z determine the coordinates (x,y) from the center of the circle ## -1.626 is the slope of line BC (simplified to 3 decimals), x and y the coordinates where it intersects with the circle ## 1.1188 is the slope of line AB and I renamed the x and y axis of this plane a and o to differenciate it from the other ones, since here x and y (thus a and o) will determinate the intersection point of the circle with line AB and will therefore have different values from the other x and y. ## the equations involving h and z state that the segment line between the center of the circle and the intersection with both AB and BC have to equal h. Why h? Because these segments are 2 of the radius and h is the y of the coordinates of the center of the circle, which is therefore the lenght of its 3rd radius. Therefore I am assigning to the equations that all three radius must be equal ## the answer to the problem is therefore the value of h, which is of course the same one as in the video

@markstahl1464 Жыл бұрын

This is a really awesome question! Thanks for sharing! I was thinking though, that it would be cool knowing the level of math required to solve the puzzle going in, that way I wouldn’t have to beat my head against the wall, not realizing that I needed something that I vaguely remember seeing maybe one time, 25 years ago. Even still, I did get a fair way through the process before getting stuck enough that I realized I didn’t have the necessary knowledge to progress. It’s kind of like playing an old school adventure game where the game state has become unwinnable without you realizing it.

@michellethaler1832 2 жыл бұрын

The trig solution is so elegant… but I’m a geometry fan- hard work, step by step .. old school . Beautiful nonetheless!

@rakesh01778 2 жыл бұрын

Nice bro keep making.

@pratapjadhao388 2 жыл бұрын

Excellent for me , Thanks sir.

@SQRTime 2 жыл бұрын

Hi Pratap. If you are interested in math competitions, please consider kzfaq.info/get/bejne/gd1ndsZ9yKm5lWg.html and other videos in the Olympiad playlist. Hope it will be useful 👍

@jimmykitty 2 жыл бұрын

Thank you boss 🥺🥰 I love your videos a lot ❤🧡

@JLvatron 2 жыл бұрын

I used the Law of Cos. Great vid-jo!

@dhruvsingla6533 2 жыл бұрын

This question comes in board examinations of class 10 and trigonometric method is very useful and time saving

@kolappan6517 Жыл бұрын

Why can't we do like this Join OM, Formed MONA is a quadrilateral Angle ONC and ANGLE OMB = 90° ( radius are perpendicular to the tangent at point of contact ) Angle O + angle A + Angle ONC + ANGLE OMB = 360° Hence, Angle O + angle A + 90° + 90° = 360° Angle O + angle A = 180 Now let us construct a perpendicular bisector OA, which divides the angle equally Hence, angle NOA = angle OAN = 45° Using trigonometry functions, Tan(theta) = opposite/adjacent Tan 45°= x/9 1 = x/9 × = 9cm That is radius = 9cm I am just a class 10 underling, so please correct me if I am wrong 😊

@Qermaq 2 жыл бұрын

5:00 or so - a far easier way for me to uses Heron's Law is to work with the prime factors. In this case, S is 2^2*3, S-7 is 5, S-8 is 2^2 and s-9 is 3. So we need the square root of the product 2^4*3^2*5. All the prime factors to an even power will form the integer portion of the answer just by dividing all the exponents by 2. The one to an odd power, 5, is left under the radical. So 2^2*3*sqrt(5) or 12sqrt(5) is the answer. (BTW if you have a prime to the third power, just split it. So if you found 7^3, the &^2 part comes out as a 7, and one 7 remains under the radical.)

@masterbullets3105 2 жыл бұрын

Just confusing problem but awesome explaination 🔥

@nevyns9285 2 жыл бұрын

i would not have thought to use the areas to figure that out, I prefer trig. very interesting way of doing it.

@williamweatherall8333 2 жыл бұрын

I'm back after trying this for a week. lol you got me good with this one. I'm interested in doing this without brute forcing it using trig.

@AzaabcG Жыл бұрын

amazing!

@fsyi8395 2 жыл бұрын

it doesn't necessary to find MB and NC, the three triangles ONC, OCB, and OBM can combine together and form a 7×r rectangle, while triangle OMA and ONA can form a 12×r rectangle

@zdrastvutye 2 жыл бұрын

two lines that are tangents have an angle together, whose bisector "leads" towards the circle's center point, there you are. after this thoughts, the whole calculation leads to 2 linear equations with 2 unknown dimensions xm and ym. einfach die winkelhalbierenden der tangenten bilden und schneiden, dann erhält man die koordinaten des Mittelpunkts

@YTN137 2 жыл бұрын

I loved the first way using to solve this problem .

@lindseyd.2707 2 жыл бұрын

trig really is overpowered

@donasaloum9154 2 жыл бұрын

Presh😍😍❤️‍🔥

@user-xt8ky2nu4u 2 жыл бұрын

I respect you !!!! I LOVE YOU !!!!

@mathewpv681 2 жыл бұрын

Very interesting problem. Let the sides of the triangle ABC be a, b, c as in standard notation. Let the circle be such that c is tangent and extended b, and extended a be also tangents. Then it can be shown that radius r=[ABC] /(S-c) where S is the semi perimeter of triangle ABC. That is r=sqrt[S×(S-a)×(S-b)÷(S-c)]

@shrijagadish 2 жыл бұрын

Very nice observation. This can be symmetrically used to find radii of circles touching one of the other sides a or b also. Further an interesting thing is that the point where the circle touches the extended side is always at a distance of S (half perimeter ) from the opposite vertex. This is true for all the three possible circles.

@mathewpv681 2 жыл бұрын

@@shrijagadish This also means that all triangles drawn by joining the two tangents from an external point to a circle such that it is also tangent to the circle will have the same perimeter, provided this tangent is on the same side of the circle as the external point.

@jacklogansmith7646 2 жыл бұрын

This was fun. I had most of the building blocks for the Geometry solution, but despite knowing that AM=AN I did not connect the dots to realize that BM=BD and CN=CD as well.

@falldamage2008 Жыл бұрын

Do you now a topic called Incentre-Excentre configuration? It covers many patterns in this diagram

@kshitijjain9302 2 жыл бұрын

Actually this is a excircle, and we have well known formulae for exradius, r=∆/(s-a)

@AmanKumar-vd1jc 2 жыл бұрын

Take A as Origin(0,0),Line AC as x axis Then, Coordinates of C is (-9,0) Using Some Trigonometry Coordinates of B is (-16/3,8√5/3) Now Radius of Circle is The Y cordinate of excentre Y cordinate of Excentre which is [(-7×0+ 9×8√5/3 +8×0)/(-7+8+9)] =24√5/10=12√5/5

@siddhantkumar1935 2 жыл бұрын

U can also use AM = 1/2(AB+AC+BC) for finding BD and DC

@mutsisloll 2 жыл бұрын

I got an A from my geometry exam, thanks to you

@RenatoSilva-sy9tj Жыл бұрын

You could use the easy relation for the radius of the circunferente related to the side 7 as r1, radius to side 8 as r2, radius to side 9 as r3: S=r1(p-7)=r2(p-8)=r3(p-9)=sqrt((p(p-7)(p-8)(p-9))

@fried_ady6318 Жыл бұрын

Hey, you mentioned the tangent half angle formula, Heron's formula, and Al Kashi's law of cosines. I was wondering if you could make a book about these lesser known formulas (I'm an accelerating yr 8). Thanks

@Gaurav_629 Жыл бұрын

Agreed

@anwardiggs8748 2 жыл бұрын

Where was this during my sophomore year? 😭

@anujsangwan6718 2 жыл бұрын

Dude that trigo solution ❤️

@SQRTime 2 жыл бұрын

Hi Anuj. If you are interested in math competitions, please consider kzfaq.info/get/bejne/gd1ndsZ9yKm5lWg.html and other videos in the Olympiad playlist. Hope it will be useful 👍

@fangtooth-1125 5 ай бұрын

There is a really easy way to do this. Circle theorems suggest that triangle ONC and ODC are congruent right angled triangles. This is the same for triangle OMB and triangle ODB. Due to this fact, we can say that NC = CD and MB = DB. Let NC = x and MB = y Then we can say that x + y = 7 Triangle ONA and OMA are congruent due to circle theorems, therefore 9 + x = 8 + y. Rearrange to get y = x+1, and solve simultaneously with x + y = 7 to get x = 3. Therefore, NA = 9+3=12. (…) means that i rounded. Also use cosine rule on triangle ABC to get angle BAC = 48.2(…). Angle OAN = 0.5angleBAC due to the two triangles being congruent as mentioned before. Using SOHCAHTOA, sin(24.1(…))=r/12 Hence, r = 12sin(24.1(…)) = 5.367

@hans-rudigerdrzimmermann 2 жыл бұрын

Nice g ood problem. I solved with the trigonometry concept.

@scottsmith2483 2 жыл бұрын

In the second solution with trigonometry, I don't see how you can say with certainty that the line segment AO must divide the angle at A into 2 equal angles theta?

@nikolakosanovic9931 2 жыл бұрын

Can you say what you need to know to solve problem, I always try to solve it but I always think I need trigonometry

@jamessanchez3032 2 жыл бұрын

What I did was extend the lines that start at point A a bit and then connect the two lines, such that you've formed a larger triangle and the circle fits in a trapezoid, with BC on "top," and the new third side of the triangle as the parallel base. I'll call the new vertices of the triangle P and Q. You have enough information to solve for PQ, PB, and QC, because 1) the big triangle is proportional to the small triangle, and 2) you have two new pairs of external tangents. PQ is 16.8, PB is 11.2, and QC is 12.6. Then, you can do some algebra recognizing that the trapezoid consists of a rectangle that is 7 x 2r, and two triangles also with height of 2r. (2r)^2 + x^2 = 11.2^2; (2r)^2 + ((16.8-7) - x)^2 = 12.6^2.

@ARN48411 Жыл бұрын

@Hanyamanusiabiasa 2 жыл бұрын

That's it, my weakness is at geometry.

@bablubhaiya1604 2 жыл бұрын

why you don't consider this formula?? r=area of triangle/(semiperimeter -length of side touch with it)

@charlesbromberick4247 2 жыл бұрын

in north america we call that formula "the crapola formula"

@Tusharsharma-dr8qh 2 жыл бұрын

This ques can also done by circle chapter theorem which is in class 10. Perimeter of triangle is equal to the sum of length of two tangent and then by trigometry various method are there this way is very complex in this video

@WahranRai 2 жыл бұрын

Let S area of triangle ABC and half perimeter p = 0.5(a+b+c) with a = 7, b=9 and c= 8 ---> we have the classical formula : Heron formula : *S = sqrt(p(p-a)(p-b)(p-c))* the external circle is the circle exscribed of the triangle ABC ---> *radius = S/(p-a)* I let substitue and find the solution

@eugene7518 2 жыл бұрын

China had no known tradition of building lifelike sculptures before Qin's reign. According to Li Xiu Zhen a senior archeologist at the Terra Cotta army site this departure of scale and style likely occurred when when influences arrived in China specifically from ancient Greece, i.e. chariots, hairstyles. They believe the appearance may have been modeled/inspired by Greek sculptures. Hence Chinese and Greeks had contact then and they exchanged knowledge such as the Pythagorean theorem.

@PurpleDuneEfa 2 жыл бұрын

Until 2:34 my solution was exactly the same, but then I drew a line from O to A dividing an angle A in to 2 exact pieces Then I used Cosine rule to find the angle A from ABC triangle. CB² = AB² + AC² - 2 * AB * AC * cos(Angle A) cos(Angle A) = (AB² + AC² - CB²)/(2 * AB * AC) Angle A = arccos((AB² + AC² - CB²)/(2 * AB * AC)) Then found the half of angle A. Then put a formula of tan(Angle A/2) = NO/NA And you get an answer of: NO = r = NA*tan(angle A/2) One thing with this is that you need a calculator that would hold onto the memory of the angle as you need it to be exact, to get the answer. I mean you can also put everything in one formula: r = NA*tan((arccos((AB² + AC² - CB²)/(2 * AB * AC)))/2) r = 12*tan((arccos((8² + 9² - 7²)/(2 * 8 * 9)))/2)

@sudoheckbegula 2 жыл бұрын

Direct formula=area/(s-a)

@SQRTime 2 жыл бұрын

Hi Nipun. If you are interested in math competitions, please consider kzfaq.info/get/bejne/gd1ndsZ9yKm5lWg.html and other videos in the Olympiad playlist. Hope it will be useful 👍

@davidseed2939 2 жыл бұрын

If the triangle hss tangent legs a,b and cross tangent c. Then the whole area is sr where s = (a+b+c)/2 The area of the triangles in the circle is cr. The area of the exterior triangle is K=sqrt(s(s-a)(s-b)(s-c)) r=K/(s-c) r²=s(s-a)(s-b)/(s-c) Here s=12 a,b,c = 9,8,7 r²=12.3.4/5 r=12¥5/5 As an exercise choose a,b,c so that r is an integer

@indianelectron5786 2 жыл бұрын

If u consider radius is same the height of that triangle, then 4.5**2 + x**2 = 7**2, which solves to sqrt(115/4) = 5.36.

@rome8726 2 жыл бұрын

Nice

@jeffeloso 2 жыл бұрын

Would be nice to create a general formula using the trig method.

@efi3825 2 жыл бұрын

If you follow along the video and replace the given sidelengths with general sidelengths a,b and c, you will end up with r = A/(s-c) (but this time, A is the area of the triangle.)

@markstahl1464 Жыл бұрын

In what course do you learn Heron’s formula? I wasn’t familiar with it.

@SuperYoonHo Жыл бұрын

thanks

@smchoi9948 Жыл бұрын

For the trigo. way, the host applied cosine law to ∠BAC to find r. Applying the law to either of the other 2 ∠s of △ABC instead could also work: Let ∠NOC = x. tan x = 3/r sec² x = 1 + (3/r)² = (r² + 9)/r² cos² x = r²/(r² + 9) As ∠BCA = 2x, by cosine law, cos 2x = (7²+9²-8²)/[(2)(7)(9)] = 11/21. As cos 2x = 2 cos² x - 1, cos² x = 16/21. What to solve is r²/(r² + 9) = 16/21. Alternatively, let ∠MOB = y. tan y = 4/r sec² x = 1 + (4/r)² = (r² + 16)/r² cos² x = r²/(r² + 16) As ∠CBA = 2y, by cosine law, cos 2x = (7²+8²-9²)/[(2)(7)(8)] = 2/7. As cos 2x = 2 cos² x - 1, cos² x = 9/14. What to solve is r²/(r² + 16) = 9/14.

@adityanarayansingh6346 2 жыл бұрын

Hi I am Aditya from India yesterday I go through ioqm aka indian olympaid qualifier in maths and all questions are very interesting . CAN YOU TELL ME ANSWERS OF SOME OF THEM? AND ALSO CAN YOU TAUGHT ABOUT PIGEON HOLE PRINCIPAL.

@michaelkennedy5584 2 жыл бұрын

Did anyone consider using calculus of variations to maximize the area of the circle within the constraining three line segments? My answer was 6...I suppose that roundoff error increases hyperbolically as the differential approaches zero.

@jomertomale 2 жыл бұрын

Circumcircle R = abc/4A Incircle r = A/s Excircles r [a/b/c] = A/(s - [a/b/c])

@Kakarot_Karthi 2 жыл бұрын

This makes you simple..... Ex radius r = (tri area)/(semi perimeter - a) r = (12√5) / (12-7) = 12√5/5 = 12/√5 Find traingle area using Heron's formula a = represents the side touches the circle s = (7+8+9)/2 = 12

@ravitejakakarala7858 2 жыл бұрын

Excircle opposite to A is r1 = delta /(s-a)

@amodbarnwal7189 2 жыл бұрын

I am from India studied in class 9 and I solve this question in 5 minutes. Using Pytagorous theorem Some circle theorem Heron formula Area of triangle 😊

@VinodKumar-rh7ds 2 жыл бұрын

Sir we can solve by the formula escribed circle

@turel528 2 жыл бұрын

Couldn't solve it in my mind. But it is a fascinating question!

@anandk9220 2 жыл бұрын

THIS IS ABSOLUTE CAKEWALK IN COMPARISON TO THE HARDEST EASY PROBLEM I SOLVED. Just TOO EASY !!! Why avoid cosine rule if we know it ???? I solved this by first finding 3 and 4 as parts of tangent segment of length 7 ( By assuming one part as x and other to be 7 - x and then using equal tangent property to solve 9 + x = 8 + 7 - x ) Then used cosine rule to find angle between sides 7 and 8, and hence found half of its exterior angle. And then simply applied trigonometric tangent ratio definition to get Radius = 5.36656312 units