My solution, using coordinate geometry: Let A be located at origin (0,0). Let O be located along the positive x-axis. Since M and N are points of tangency, then OM⊥AM and ON⊥AN. So triangles AOM and AON are both right triangles that share hypotenuse AO and have congruent legs (OM=ON=r). Therefore they are congruent triangles by HL (hypotenuse-leg). Let ∠OAM = ∠OAN = θ Since AB = 8, then B = (8 cos θ, 8 sin θ) Since AC = 9, then C = (9 cos θ, −9 sin θ) Since BC = 7, then BC² = 7² (9 cos θ − 8 cos θ)² + (−9 sin θ − 8 sin θ)² = 49 (cos θ)² + (−17 sin θ)² = 49 cos²θ + 289 sin²θ = 49 cos²θ + sin²θ + 288 sin²θ = 49 1 + 288 sin²θ = 49 288 sin²θ = 48 sin²θ = 48/288 = 1/6 → sin θ = 1/√6 cos θ = 1 − 1/6 = 5/6 → cos θ = √5/√6 tan θ = sin θ / cos θ = 1/√5 In △AMO: tan θ = OM / AM 1/√5 = r / 12 r = 12/√5

Good problem and solution, but you could have saved some time. BC is the base of OBC and we already know that's it's 7, and being a tangent, h=r. Then you just have to solve 12√5 + ½7r = ½8r + ½9r = ½17r 12√5 = 5r Still, nice problem and approach overall. 👍

You can use law of cosines for the triangle ABC to find cosA and then cos(A/2),after that you find the length of OA and then youbuse pythagorean theorem on OAM to find OM which is the radius of the circle

AM and AN have equal lengths, so 9 + x = 8 + 7 -x. Solving for x, x = 3. AM and AN, therefore, each have length 12. ΔAMO and ΔANO have bases of length 12 and heights of the circle's radius, call that r, so each Δ has area (1/2)(12)r = 6r and combined area 12r. ΔBOC has base 7 and height r, so area = (1/2)(7)r = (3.5)r. Area of ΔBMO = area of ΔBPO and area of ΔCNO = area of ΔCPO, so combined area of ΔBMO and ΔCNO equals combined area of ΔBPO and ΔCPO equals area of ΔBOC. So area of pentagon BMONC = twice the area of ΔBOC equals 7r. Deduct the area of the pentagon from combined area of ΔAMO and ΔANO to get the area of ΔABC, so 12r - 7r = 5r is equal to the area of ΔABC. However, Math Booster computed that area as 12√5. So 5r = 12√5 and r = 12/(√5).

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

Let O be the center of the circle and P be the point of tangency between BC and circle O. If CP = x, then BP = 7-x. As B and C are points intersected by two tangents, the lengths of those line segments are identical for each point. So BM = BP = 7-x and CN = CP = x. Similarly, as A is also a point intersected by two tangents, the lengths of those line segments are identical. So: AM = AN AB + BM = AC + CN 8 + (7 - x) = 9 + x 15 - x = 9 + x 2x = 15 - 9 = 6 x = 6/2 = 3 So CP = CN = 3, BP = BM = 7-3 = 4, and AM = AN = 9+3 = 12. Let ∠BAC = 2α. By the law of cosines: cos(2α) = (AB²+AC²-BC²)/2(AB)AC cos(2α) = (8²+9²-7²)/2(8)9 cos(2α) = (64+81-49)/144 cos(2α) = 96/144 = 2/3 Draw OA, OM, and ON. As AM and AN are tangent to circle O, ∠OMA = ∠ANO = 90°. As OM = ON = r, being radii of circle O, AM = AN, and OA is shared, right triangles ∆OMA and ∆ANO are congruent, and thus OA bisects 2α. cos(2α) = 1 - 2sin²(α) 2/3 = 1 - 2sin²(α) 2sin²(α) = 1 - 2/3 = 1/3 sin²(α) = (1/2)(1/3) = 1/6 sin(α) = √(1/6) = 1/√6

I got to this answer by a very different and long-winded method. I set up a coordinate system with A as the origin and AO as the x axis (the circle centre is O.) AO bisects BAC, with half angle 'a'. I then gave the coords of B and C in terms of cos(a), sin(a), and from these point P. O is at (z,0) where z^2 = r^2 + 12^2, and OP^2 = r^2. The evaluation of OP^2 eventually reduced to a simple equation in c = cos(a): c^2 = 5/6, but c^2 = 12^2/(r^2 + 12^2), therefore 5r^2 = 12^2, r = 12/sqr(5). Whew!

An alternative is to scale up triangle ABC by an unknown factor K to the point you have the circle inscribed in an inflated triangle AB’C’. The side that corresponds to BC, length 7, is B’C’ which has length 7K= 9K-12 +8K-12. This is from the tangencies from vertices B’ and C’. K is 12/5. The semi perimeter of AB’C’ is K times bigger than for ABC, [AB’C’] is K^2 times [ABC]. From Area-perimeter-radius formula get radius of inscribed circle as Ksqrt5 which is 12/sqrt5.

Good problem and solution, but there's a much faster solution. First, we see BM = CN+1 and since BM=7-x and CN=x, then x=3 and 7-x=4. This makes AM = AN = 12. Then we solve BAC by the cosine rule: 7^2 = 8^2 + 9^2 - 2*8*9*cos (BAC) --> 49 = 64+81-144*cos (BAC) = 145 - 144 cos (BAC) --> 96=144*cos (BAC) --> cos (BAC) = 2/3. Then we look at AOM triangle (or AON same thing). R/ sin (MAO) = 12/ sin (MOA). We know MAO = ABC/2 --> cos (MAO) = sqrt [(1+ cos (BAC))/2] = sqrt [(1+2/3)/2] = sqrt (5/6) --> sin (MAO) = sqrt (1/6) --> sin (MOA) = cos (MAO) = sqrt(5/6) --> R = 12 * sin (MAO) / sin (MOA) = 12 * sqrt (1/6) / sqrt (5/6) = 12 / sqrt (5) = 12 * sqrt (5) / 5.

Use Cosine law to determine Cos(A) = 2/3 Use half angle formula, you can determine COS(A?2) = square root of 5/6). Then Tan(A/2) can be determned to be square root of 1/5. Finally Tan(A/2/ = square root 1/5) = radirus r/12. ===> Radius = 12/(square root of 5)

You can early in the calculations get the value for X. (7-X) + 8 = X + 9. X = 3, (7 - X) = 4.

Having established BM=BP=4, CN=CP=3, AM=AN=12 Use t = r tan(theta) where t is the length of a tangent, r is radius of the circle, and theta is the angle subtended by the tangent at the circle centre Let angle MOB = theta, NOC = phi so MOA = theta+phi r tan(theta) = 4 r tan(phi) = 3 r tan(theta + phi) = 12 r (tan(theta) + tan(phi))/(1 - tan(theta)tan(phi)) = 12 12/r = (4/r + 3/r)/(1 - 12/r^2)) 12 - 144/r^2 = 7 144/r^2 = 5 r^2 = 144/5 r = 12/sqrt(5)

I used law of cosines to find angle A. Solved for x = 3. Used angle A/2 and length AN = 12 to solve for radius. Fairly easy if trig is allowed. Nobody said it wasn't.

Nice

After you found that [ABC]=12√5, we can proceed in this way. AM=AN --> 8+7-x=9+x --> 2x=6 --> x=3 --> AM=AN=12 Triangles BMO = BPO and triangles PCO=CNO and triangles AMO=ANO, so we can write: [AMON] = [AMO]+[ANO] = [ABC]+[BMO]+[BPO]+[PCO]+[CNO]--> 2[AMO]=[ABC]+2[BMO]+2[CNO] --> AM*MO = 12√5+BM*MO+CN*MO--> MO(AM-BM-CN)=12√5 --> MO(12-4-3)=12√5 --> MO=12√5/5

A convenient way to determine the radius is to consider the the circle as excircle of the triangle. Then r= A/(s-7), A =area ,s= half circumference . This formula is not difficult to derive.

From this you can also get: r = [ABC] / (S - bc) (bc is the side that is not extended)

Use tangent of the sum: 5:00 ∠O+∠A=pi, half of it is ∠BOC+1/2*∠A=pi/2, and ∠BOC=∠BOP+∠COP By the lengths we know tan∠BOP=4/r, tan∠COP=3/r and tan1/2*∠A=r/12 tan∠BOC=tan(∠BOP+∠COP)=tan(pi/2-1/2*∠A)=12/r by the sum of tangent formula, tan(∠BOP+∠COP)=(4/r+3/r)/(1-12/r^2)=tan(pi/2-1/2*∠A)=12/r and solve it, r=12/√5

Like the equation of triangle within circle.

Very good! I love it!

Which tool are you using to write on the whiteboard for teaching?

good problem😮

Respuesta es R = 4,83

R=6

Nice one.....

Using little trigonometry: (BC)^2=(AB)^2+(AC)^2-2(AB)(AC)cosA, so solving for cosA we get cosA=2/3, so sinA=sqrt(5)/3, therefore tanA=sqrt(5)/2 and because tanA=2t/(1-t^2), where t=tan(A/2), solving for t we get t=1/sqrt(5). As explained by others, CN=x=3, so AN=12, therefore ON=r=(AN)tan(A/2)=12/sqrt(5)