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 😍😍
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)
@@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 !!!!
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.
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.
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..
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)
At 0:29, MB=BD=x say................................(1) tangent rule. DC=(7-x)=NC.................................(2) tangent rule. MA=x+8=NA=(7-x)+9...................(3) tangent rule. x+8=(7-x)+9 >>>>>>x=4..............(4) We need angle BAC, Take Cosine Rule about triangle BAC, 7^2=8^2+9^2-2*8*9*cosBAC 49 =64+81-144cosBAC cosBAC =2/3 angle BAC=arcos(2/3)..................(5) We need length MN through circle, Take Cosine Rule about triangle MAN, MA=4+8=12 and AN=3+9=12 MN^2=12^2+12^2-2*12*12*(2/3) =288/3=96....................................(6) Take Cosine Rule about OMN, Cosine angle NOM= -cos BAC, MN^2=r^2+r^2-2*r^2(-cosBAC)=2r^2(1+2/3) 96=2*r^2*(5/3) 48=(5/3)*r^2 r^2=(144/5) r=12/(sqrt5) units in length. we need the +ve root. That is our answer.
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
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.
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.
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
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
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 😊
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
~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.
Hi Bugrick. If you are interested in math competitions, please consider ruclips.net/video/Kw7CcLcCUe4/видео.html and other videos in the Olympiad playlist. Hope it will be useful 👍
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
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.
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.)
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
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
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)
Area of triangle given 3 sides: Using Heron's Formula: A=√(s(s-a)(s-b)(s-c)) to find the semi-perimeter S S=(7+8+9)/2 =12 A=√(12(12-7)(12-8)(12-9)) = 12√5 The area of the triangle given excircle is A=r(s-a), where a is the length of triangle in which circle is tangent to it. then, 12√5 = r(12-7) r=12/√5 units
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
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
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
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
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
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.
A = tangency point side 9 B = vertex sides 7 and 9 C = vertex sides 9 and 8 D = vertex sides 7 and 8 E = tangency point side 8 F = tangency point on side 7 O = center of circle S (ACEO) = quadrilateral area ACEO = 2 * (area ACO) (1) and also S (ACEO) = ABDEO area + BCD triangle area (2) Area ACO = AC * R / 2 where R is the radius of the circle From AC = CE, AB = BF, FD = DE, BF + FD = 7, AC = AB + BC = CD + DE, follows AC + CE = BF + BC + CD + FD = BC + CD + BD = 24 = 2AC ... AC = 12, AB = 12-9 = 3; DE = 12-8 = 4 area ACO = AC * R / 2 = 12 * R / 2 S (ACEO) = 2 * 12 * R / 2 = 12 * R ABDEO area = 2 * ABO area + 2 * DEO area = 3 * R / 2 * 2 + 4 * R / 2 * 2 = 7 * R Applying Heron's formula ... BCD area = √semiperimeter (= p) * (p-7) * (p-8) * (p-9) = 12 * √5 From (1) and (2) ... 12 * R = 7 * R + 12 * √5 from which R = 12/5 * √5
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.
I can tell from the following timestamps: 0:00 Problem 1:22 Geometry : Takes 4:13 to explain it 5:35 Trigonometry : Takes 2:07 to explain it (video is 7:42) The Trigonometry solution is actually easier than the Geometry solution! I can visualize how to calculate this using Trig, but not without.! So, right away, yet another problem where the invention of Trig simplifies life. I reached this conclusion by not even watching any of the solution.
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
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
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)]
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.
@@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.
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))
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.
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
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
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.
Wait .. maybe I am wrong .. isn't this set up impossible ... if you construct a line from the center of the circle to angle A .. it would bisect BC in a right angle .... that would make the triangle ABC an isosceles ...
Hi Anuj. If you are interested in math competitions, please consider ruclips.net/video/Kw7CcLcCUe4/видео.html and other videos in the Olympiad playlist. Hope it will be useful 👍
just use the formula: Area of triangle = radius of the escribed circle * (semiperimeter of the triangle - length of the tangent side) you have all the lengths of the sides of the triangle so you can get the Area of the triangle using Heron's formula; you can easily get the semi perimeter of the triangle by diving its perimeter by 2. then you have it. just solve for the missing radius of the escribed circle.
Hi Nipun. If you are interested in math competitions, please consider ruclips.net/video/Kw7CcLcCUe4/видео.html and other videos in the Olympiad playlist. Hope it will be useful 👍
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 😊
Since this is a common problem, there is a formula generated to it already Area of triangle = (s - a) × r where s is the semi perimeter 🙂🙂 and a is the length of the tangent side Wanna try Area of the triangle (Heron's) √{s(s-a)(s-b)(s-c)} where s = half the perimeter of 📐 = 12 a = 7 (tangent side) b = 8 c = 9 (could interchange) By substituting Area of 📐= 12√5 From the derived formula Area of triangle = (s - a) × r r = Area of triangle / (s - a) r = 12√5 / (12 - a) r = 12√5 / 5 This way, no matter which side you let to be tangent to the circle. it would be easy to get the radius as semi perimeter and area do not vary as long as you have the same triangle. Just the length of the tangent side a. 🙂🙂🙂
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.
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.)
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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 😍😍
Gammac mageth favourite channel ekak dana hama ganama hadnw 🖤❤️
Hi. I'm from your neighbouring country.😁😁
@@user-ci5kh7pl8y
India 🇮🇳 ✋
@@user-ci5kh7pl8y Me too neighbouring country 😃
India
@DoTheMath lmaooo
Dark Mode: Activated. Nice. :)
now its just organic chemistry channel
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)
That looks far more complicated than avoiding the law of cosines.
@@Tiqerboy Oh, I thought it brought out the symmetries in the problem. Thnx for comment
No, Just use similiarity to avoid complexity
@@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 !!!!
@@philosophical_aspect. What are l and b?
Its taught in india in class 10/11 th.
So it was easy for me.
Its basically ∆/s-a
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
But obviously you didn't know how to derive it hahaha
@@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 ,
If you want just check class 10 cbse circle previous year questions , you will find it.
@@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
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.
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.
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..
I'm from India, 17 years old and I solved this question by my own made Formula within 5 seconds.
What is it?😊
@@shaaktam3344 don't bother
I have seen of
Tons of people like him
If he really made such a formula, he should have been famous by how
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)
At 0:29,
MB=BD=x say................................(1) tangent rule.
DC=(7-x)=NC.................................(2) tangent rule.
MA=x+8=NA=(7-x)+9...................(3) tangent rule.
x+8=(7-x)+9 >>>>>>x=4..............(4)
We need angle BAC,
Take Cosine Rule about triangle BAC,
7^2=8^2+9^2-2*8*9*cosBAC
49 =64+81-144cosBAC
cosBAC =2/3
angle BAC=arcos(2/3)..................(5)
We need length MN through circle,
Take Cosine Rule about triangle MAN,
MA=4+8=12 and AN=3+9=12
MN^2=12^2+12^2-2*12*12*(2/3)
=288/3=96....................................(6)
Take Cosine Rule about OMN,
Cosine angle NOM= -cos BAC,
MN^2=r^2+r^2-2*r^2(-cosBAC)=2r^2(1+2/3)
96=2*r^2*(5/3)
48=(5/3)*r^2
r^2=(144/5)
r=12/(sqrt5) units in length. we need the +ve root.
That is our answer.
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
Where did u get the s=12? U should've explained that, too.
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.
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.
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
R= area of traingle/ semiperimeter - side touched by Circle
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
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 😊
تمرين جميل جيد . شرح جيد واضح مرتب . شكرا جزيلا لكم استاذنا الفاضل والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .
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
~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.
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
escribed circles : r1 = A/( s-a) where A = rt { s(s-a)(s-b)(s-c)} and s = (a+b+c)/2
This is the creative logic
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😂
@@newzero1000 exactly
That's right
6:00 how this triangle could be possible? 4/3 should be equal to 8/9 then
In Darkmode now, nice! Interessting puzzle :)
Hi Bugrick. If you are interested in math competitions, please consider
ruclips.net/video/Kw7CcLcCUe4/видео.html and other videos in the Olympiad playlist. Hope it will be useful 👍
Your way of explanation is great. Love from INDIA.
01-03-2022..1:58 PM
Nice question . Please give these types of question countinously. It refrains our mind. It took me more than five minutes to solve.
Welcome back, Presh.
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
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.
I so much like your trigonometry method .......it's so easier than geometry
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.)
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
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
Thank you, can't believe I went my whole life never having heard of that so-easy Heron formula oO
Thank you! The black screen is best for my eyes.
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)
Area of triangle given 3 sides: Using Heron's Formula: A=√(s(s-a)(s-b)(s-c))
to find the semi-perimeter S
S=(7+8+9)/2 =12
A=√(12(12-7)(12-8)(12-9)) = 12√5
The area of the triangle given excircle is A=r(s-a), where a is the length of triangle in which circle is tangent to it. then,
12√5 = r(12-7)
r=12/√5 units
thanks for boosting my love for maths!
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
My maths olympiad exam is on 6th march love from india
Who knew thanos was Indian
Thanoswaram iyer
Ioqm
@@qwertyuiop-pc2xw
Thanoswamy muttuswammy iyyer.
@@Anonymous-kw7ls thanoswamy mutthuswamy venugopal iyyer
Loved the new dark theme ❤️
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
Sweet, merciful dark theme.
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
I used the Law of Cos. Great vid-jo!
The geometry portion was elegant. Good work.
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
Thank you!
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
Excellent for you to this 👏question
Nice question👌🏼
Circumcircle R = abc/4A
Incircle r = A/s
Excircles r [a/b/c] = A/(s - [a/b/c])
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
The trig solution is so elegant… but I’m a geometry fan- hard work, step by step .. old school . Beautiful nonetheless!
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.
why you don't consider this formula??
r=area of triangle/(semiperimeter -length of side touch with it)
in north america we call that formula "the crapola formula"
A = tangency point side 9
B = vertex sides 7 and 9
C = vertex sides 9 and 8
D = vertex sides 7 and 8
E = tangency point side 8
F = tangency point on side 7
O = center of circle
S (ACEO) = quadrilateral area ACEO = 2 * (area ACO) (1) and also
S (ACEO) = ABDEO area + BCD triangle area (2)
Area ACO = AC * R / 2 where R is the radius of the circle
From AC = CE, AB = BF, FD = DE, BF + FD = 7, AC = AB + BC = CD + DE,
follows AC + CE = BF + BC + CD + FD = BC + CD + BD = 24 = 2AC ...
AC = 12, AB = 12-9 = 3; DE = 12-8 = 4
area ACO = AC * R / 2 = 12 * R / 2
S (ACEO) = 2 * 12 * R / 2 = 12 * R
ABDEO area = 2 * ABO area + 2 * DEO area = 3 * R / 2 * 2 + 4 * R / 2 * 2 = 7 * R
Applying Heron's formula ...
BCD area = √semiperimeter (= p) * (p-7) * (p-8) * (p-9) = 12 * √5
From (1) and (2) ... 12 * R = 7 * R + 12 * √5 from which R = 12/5 * √5
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.
I can tell from the following timestamps:
0:00 Problem
1:22 Geometry : Takes 4:13 to explain it
5:35 Trigonometry : Takes 2:07 to explain it (video is 7:42)
The Trigonometry solution is actually easier than the Geometry solution! I can visualize how to calculate this using Trig, but not without.!
So, right away, yet another problem where the invention of Trig simplifies life.
I reached this conclusion by not even watching any of the solution.
Me who dont understand most of those greek symbols: geometry is logic while the other one is alien language
@@AlwaysOnForever γεια σου
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
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
cos(2𝜃) = 2 ∕ 3
Using the double angle formula, we get
2 cos²(𝜃) − 1 = 2 ∕ 3 ⇒ cos²(𝜃) = 5 ∕ 6
From the trigonometric 1, we then get
sin²(𝜃) = 1 − cos²(𝜃) = 1 ∕ 6
Thus, tan²(𝜃) = sin²(𝜃) ∕ cos²(𝜃) = 1 ∕ 5
0° < 𝜃 < 90° ⇒ tan(𝜃) = 1 ∕ √5
From the diagram we have tan(𝜃) = 𝑟 ∕ 12,
which gives us 𝑟 = 12 tan(𝜃) = 12 ∕ √5 = 12√5 ∕ 5
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)]
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.
@@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.
one could also proceed by first finding the radius of the incirclr of triangle ABC which is root 5. and then use Basic Proportionality theorem.
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.
Sir we can solve by the formula escribed circle
Presh😍😍❤️🔥
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))
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.
Do you now a topic called Incentre-Excentre configuration? It covers many patterns in this diagram
U can also use
AM = 1/2(AB+AC+BC)
for finding BD and DC
trig really is overpowered
Actually this is a excircle, and we have well known formulae for exradius, r=∆/(s-a)
Can you say what you need to know to solve problem, I always try to solve it but I always think I need trigonometry
Nice g ood problem. I solved with the trigonometry concept.
06:55 AN = 12 because NC = 3 was known from the geometry solution. I stopped at tan(theta) = r / (9 + NC) because I didn't have a value for NC.
Other way to get 12 is by taking the half of the perimeter of the triangle so you get the length of the tangents
8:08 we allready know total are and other part just subract ops
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
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
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.
i would not have thought to use the areas to figure that out, I prefer trig. very interesting way of doing it.
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.
At 6:00 it is angular bisector but
4/3 is not equal to 8/9 can anyone plzz tell ??
It's because 4 & 3 are the lengths of BD & CD, but D is not at the point of intersection shown (where OA bisects BC).
Wait .. maybe I am wrong .. isn't this set up impossible ... if you construct a line from the center of the circle to angle A .. it would bisect BC in a right angle .... that would make the triangle ABC an isosceles ...
Dude that trigo solution ❤️
Hi Anuj. If you are interested in math competitions, please consider
ruclips.net/video/Kw7CcLcCUe4/видео.html and other videos in the Olympiad playlist. Hope it will be useful 👍
Love these solutions ❤
just use the formula: Area of triangle = radius of the escribed circle * (semiperimeter of the triangle - length of the tangent side)
you have all the lengths of the sides of the triangle so you can get the Area of the triangle using Heron's formula; you can easily get the semi perimeter of the triangle by diving its perimeter by 2. then you have it. just solve for the missing radius of the escribed circle.
The two ways are amazing. Thank you!
Sir I think you can go back to the white background, the black one is not doing gud
I got an A from my geometry exam, thanks to you
That's it, my weakness is at geometry.
Direct formula=area/(s-a)
Hi Nipun. If you are interested in math competitions, please consider
ruclips.net/video/Kw7CcLcCUe4/видео.html and other videos in the Olympiad playlist. Hope it will be useful 👍
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
😊
I loved the first way using to solve this problem .
Since this is a common problem, there is a formula generated to it already
Area of triangle = (s - a) × r
where s is the semi perimeter 🙂🙂 and a is the length of the tangent side
Wanna try
Area of the triangle (Heron's)
√{s(s-a)(s-b)(s-c)}
where s = half the perimeter of 📐 = 12
a = 7 (tangent side)
b = 8 c = 9 (could interchange)
By substituting
Area of 📐= 12√5
From the derived formula
Area of triangle = (s - a) × r
r = Area of triangle / (s - a)
r = 12√5 / (12 - a)
r = 12√5 / 5
This way, no matter which side you let to be tangent to the circle. it would be easy to get the radius as semi perimeter and area do not vary as long as you have the same triangle. Just the length of the tangent side a. 🙂🙂🙂
Best place to find quality math puzzle.
In fact, this is a basic problem in Olympiad. s=12, therefore, if we write Heron’s formula here, the root 12*4*3*5=(12-7)r
So r=12root5/5
I constructed it on AutoCAD and got the same answer. Tanzania
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.
..
Where was this during my sophomore year? 😭
these type of Circle are called e circle and is equal to A/s-a where A is area of triangle and a is t
side opposite to circle
hi, on 5:43 how can we be sure the angles are equal? pls someone explain
We can draw radius OM, and we can prove triangle ONA and OMA as congruent, and then use CPCT
Would be nice to create a general formula using the trig method.
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.)
I used Pythagoras theorem in ∆AMO and ∆ABD
But I got radius= 12/√3
AD= √8²-4² = 4√3
Now, in ∆AMO ,
r² + AM² = AO²
r² + 144 = {r + (4√3)}²
r² +144 = r² + 48 + 8√3r
96=8√3r
r= 12/√3
∆ABD is not a right triangle. If it were, BD = CD
@@gopanneyyar9379 he is right bro i have applied trigo in it and i also got 4root 3
@@aakarshjha1633 Correct answer is 12/√5. Not 12/√3 which is equal to 4√3. He got wrong answer because he assumed that ∆ABD is a right triangle.