Math Olympiad Challenge | Find Radius of the Circle | Important Geometry Skills Explained

Find Radius of the Circle | Math Olympiad Challenge | Important Geometry Skills Explained

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[dantallman]

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.

[Ni]

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. 👍

[gibbogle]

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!

[vaggelissmyrniotis]

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

[MarieAnne.]

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

[quigonkenny]

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 <== 0° < α < 90°

cos²(α) + sin²(α) = 1
cos²(α) + 1/6 = 1
cos²(α) = 1 - 1/6 = 5/6
cos(α) = √(5/6) = √5/√6

tan(α) = sin(α)/cos(α)
tan(α) = (1/√6)/(√5/√6)
tan(α) = 1/√5

Triangle ∆OMA:
tan(α) = OM/AM
1/√5 = r/12
r = 12/√5 ≈ 5.367 units

[juanalfaro]

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.

[اممدنحمظ]

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

[jimlocke]

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).

[chingweilai]

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)

[bpark]

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

[randyrogers]

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.

[pwmiles]

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)

[renesperb]

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.

[EnnioPiovesan]

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] =
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

[xwyl]

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

by the sum of tangent formula,
and solve it, r=12/√5

[meowsnipurr]

From this you can also get:
r = [ABC] / (S - bc)

(bc is the side that is not extended)

[modern_genghis_khan]

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

[wira]

Like the equation of triangle within circle.

[stathislourantos]

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)