# Challenge 65: Tangent Circles and Angle Chasing

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• Published on Nov 1, 2018
• Congratulations to Vishaal Ram, reynolds45, Minh Cong Nguyen, enisheadpay, Kevin Tong, Bigg Barbarian, Rishav Gupta, Nicola C, and Czeckie for successfully solving the last week's math challenge question! Vishaal Ram was the first person to solve the question.
Your support is truly a huge encouragement.
Every subscriber and every like are wholeheartedly appreciated.
Welcome, everyone! My channel hosts one weekly math challenge question per week (made by either myself, my family, or my friends), which will be posted every Wednesday. Please comment your proposed answer and explanation below! If you are among the first ten people with the correct answer, you will be recognized in the next math challenge video. The solution to this question and new question will be posted next Wednesday.
For more Weekly Math Challenges:

• Sashank Sriram 10 months ago

Let me just name the circles A, B and C for convenience.
>So, we know that the radius of A and B are 3 and 5 respectively, and that angle(FDG)=30
>We need to find angle(EAD), so let's just say it equals 't'.

Now, let's focus on triangle(DFB)
>The angle in a semicircle is a right angle, which means that angle(DFB)=90
>By angle sum property, we can say that angle(FDB)=60
>DFB being a 30-60-90 triangle must bear a ratio of 1:root(3):2
>That means that DF=5root(3)

Now let's focus on triangle(FDG)
>By the radial isosceles property, angle(FDG) = angle(GED) = 30

Now let's focus on circle C
>Angle subtended at any point on the circumference by the same chord, is the same angle
>Implies angle(FDG)=angle(FEG)=30 and angle(DFG)=angle(DEG)=30

The above case implies that angle(FED)=30+30=60

Now let's focus on triangle(AED)
>Angle(AED), as the angle in a semicircle, is 90
>By exterior angle property, angle(EDB)=90+t
>We can deduce that angle(EDF)=60+t
>We also know that AD=6, so AE=6cost, DE=6sint

Now let's focus on triangle(EDF)
>By angle sum property, angle(EFD) = 60-t
>By sine rule, ED=10sin(60-t)

We see that ED=10sin(60-t)=6sint
10(sin60cost - sintcos60)=6sint
(5root(3))cost-5sint=6sint
11sint=(5root(3))cost
Squaring both sides
121sin^2 t = 75 cos^2 t
121 - 121cos^2 t = 75 cos^2 t
121 = 196 cos^2 t
cos^2 t = 121/196
cos t = +- 11/14

• Sashank Sriram 10 months ago

@Hung Hin Sun Thanks for pointing that out. I've edited my comment.

• Hung Hin Sun 10 months ago

The mistake appears in line 24, since sin(60-t) should be equal to sin60cost-cos60sint.

• Bigg Barbarian 10 months ago

11/14.
First I placed a coordinate system with its origin in D, so alpha becomes 9=(x+3)**2+y**2 and beta 25=(5-x)**2+ y**2.
Next I solved for F to be (30/4,-5/2*sqrt(3)).
Now F, D and G define gamma as 25=(5/2-x)**2+(5/2*sqrt(3)+y)**2.
Intersecting alpha and gamma yields D at (0,0) and E at(-225/98,-165*sqrt(3)/98).
Now we consider a right triangle consisting of A, E and the P the intersection of AD and the perpendicular of E on AD.
Obviously AP is 6-225/98=363/98 and its hypothenuse is sqrt((363/98)**2+(165*sqrt(2)/98)**2)=33/7,
hence cos (EAD) = (363/98)/(33/7) = 11/14.

• Sourin Chatterjee 10 months ago

Sorry of Vishall Ram is edited though you accepted breaking your rules yourself.

• LetsSolveMathProblems  10 months ago +1

The only reason I implemented the "no edit rule" was to prevent the viewers from posting a flawed answer, then changing it significantly later. As far as I am aware, editing one's answer does not reset the timestamp of your comment, which is a problem since I wish to recognize the first 10 people with correct solutions (not first 10 commenters). As stated in the video, in this particular case, I already knew Vishaal Ram was the first person to post a complete, correct solution, so I decided to accept it. Of course, in almost every other case, I cannot bypass this rule since I do not keep track of when every comment is uploaded, along with whether the explanation is correct at the time of the first upload.

• Florent Bréart 10 months ago

Let I be the intersection of γ and (AE).
Because AED is a right triangle, we have :
AGI = DGI = 180° - DEI = 180° - 90° = 90°
So AGI is a right triangle.
In AGI, cos(EAD) = AG / AI, let's calculate AI.
Also, because DGF is an isosceles triangle :
DIG = DFG = FDG = 30°
Then IG = DG / tan(30°) = DG * √(3) = 5√(3)
Hence, AI = √(AG^2 + IG^2) = √(11^2 + (5√(3))^2) = √(121 + 75) = √(196) = 14
Therefore, cos(EAD) = AG / AI = 11 / 14

• 4ajit9 10 months ago +1

• Quwertyn 10 months ago

Let's consider an inverse with center in D and radius of sqrt(10). α and β will become parallel lines we will call p and k. γ will become a third line intersecting them at E* and F* (X*=point X after the inverse. The exception will be D* and that's how we will call the inverse center). A*B* will be perpendicular to both p and k. G* will be distance 2 from D*, B* will be distance 1 from D* and A* will be distance 5/3 from D*. Angle FDG will become F*D*G* and EAD will become A*E*D*. D*G*F* will form an isosceles triangle (F*B* divides D*G* in half and is perpendicular), so A*G*E*=G*D*F*=π/6. Let's consider the triangle E*A*G*. E*G*=A*G*/cos(π/6)=11/(3*sqrt(3)/2)=22sqrt(3)/9. E*F*=E*G*-F*G*=22sqrt(3)/9-6sqrt(3)/9=16sqrt(3)/9
E*D*^2=E*F*^2+D*F*^2-2cos(E*F*D*)D*F* * E*F*=196/27
Considering the triangle A*E*D* we get A*E*^2=196/27-75/27=121/27
A*E*=11/3sqrt(3)
E*D*=sqrt(196)/3sqrt(3)=14/3sqrt(3)

• TrollTree 10 months ago

11/14 constructed in geogebra

• JivaRaj Shetty 10 months ago

Let the required angle be x. Observe triangle EAG using cosine law we arrive with side EG^2= 121-96cos^2(x) and angle EDG is 90+x.
Also, the circumradius of gamma is 5 (sine rule on the triangle FDG). Hence EG/sin(EDG)=2(5).
On further simplification, we arrive at sqrt(121-96cos^2(x))/cos(x)=10
Therefore 121=196cos^2(x) ,implies cos(x)=11/14.

• santosh tripathy 10 months ago

We Claim that the Ans is 11/14
Join DE,DF,GE,AE,GF,EF,We will work with angles in degrees rather than radians thus we omit writing degrees after the angles
Let Angle EAD=x then Angle EDA=90-x,given that Angle FDB=30,Since ADB are collinear thus Angle EDF=60+x
Angle GDF=Angle GFD =30 and since EDGF is cyclic thus Angle GEF=Angle GDF=30 and Angle DEG=Angle GFD=30
thus Angle DEF=Angle DEG+Angle GEF=60,In Triangle DEF Sum of angles=180 thus Angle DFE=60-x
Now in Triangle DGF,DF=DG=5 and angle GDF=GFD=30 thus DF=5*(Square root of 3) (By the Projection Formula)
Note That Angle AED=90 thus ED=AD(Cosx)=6Cosx
Now Using the sine Rule in Triangle EDF - ED/Sin(EFD)=DF/Sin(DEF) which implies 6Sinx/Sin(60-x)=5*(Square Root of 3)/Sin(60)
Now After Some Calculatons(Expanding Sin(A-B) and Dividing throuout by Sin(x)) We get Cotx=11/5*(Square Root of 3)
THUS COS(x)=11/14

• himanshu mallick 10 months ago +1

Hi there!
Could you grant me the permission to post your Weekly Math Challenges on AOPS? Seriously, your problems are very nice and original.
I would only post the problems whose solution video has been made.

• himanshu mallick 10 months ago

Thanks

• LetsSolveMathProblems  10 months ago +2

You are more than welcome to post or share Weekly Math Challenge on any site, as long as you are not selling the problems for money. =)

Someone always has to do it the hard way... Let D be the origin. The cosine rule lets us calculate the distance to F, which gives F = (15/2, 5 sqrt(3)/2). Then let the circle gamma be (x-a)^2 + (y-b)^2 = c^2 and solve so it passes through D, G and F, which gives a = 5/2, b = 5 srt(3)/2, c = 5. Then find the intersection of that with the circle alpha (x+3)^2 + y^2 = 9, to get E = (-225/98, 165 sqrt(3)/98). Then elementary trig gives cos(EAD) = 11/14. I'll await the elegant and geometrically insightful method next week. :)

• Ricardo Mogollon 10 months ago

at first i concentrated at circle B, because DG and FG are radius of te circle their distance is equal to five, using the fact that the triangle DGF is an isosceles triangle we can conclude that diagonal DF of cuadrilateral EDGF is equal to 5sqr(3) and angle DGF is equal to 120 degrees. taking into account that the previous cuadrilateral is cyclical, we have that opposite angles are supplementary, therefore we can see that angle DEF is equal to 60 degrees, using the external angle theorem and the fact previously found, we can see that angle EDF=S+60 and angle EFD=60-S where S is the angle EAD, using the theorem of sins on triangle EDF it can be shown that the distance ED corresponds to 5(sqr(3)*cos(s)-sin(s)). Finally using the fact that AED is a right triangle we can se that sin(s)=(5(sqr(3)*cos(s)-sin(s)))/6, solving with respect to sin(s) and using the fact that sin^2(s)+cos^2(s)=1 we arrive to our answer

• Brent Kreinop 10 months ago

11/14. Found by inspection constructing the diagram in GeoGebra classic. Pick the origin as the center of circle alpha. This makes point G be (8,0). Point F is (10.50, 4.33), point E is (0.70,2.92), all to two decimal places. The circle containing F, G and D is radius 5, centered at (5.5, 4.33). Point E is therefore the intersection of this 3 point circle with circle alpha. Drawing the segment AE, and a circle radius 1 centered at A. Mark point I as the intersection between AE and this circle. Point I is (-2.21,0.62) to two decimal places. Expanding the display to 15 decimal places yields a fraction very close to 5.5/7, or 11/14.

• Aswini Banerjee 10 months ago

No,the correct construction is like this

• LetsSolveMathProblems  10 months ago +1

Although GeoGebra is an excellent way of visualizing the problem, a solution solely based on it will not be accepted. Your answer suggests that the answer is 11/14. It does NOT prove that it is 11/14.

• Jeremy Weissmann 10 months ago

I wish I had learned about cyclic quadrilaterals! Instead I placed the origin at the center of the circle with radius 3, and calculated the coordinates of the center of the third circle as (5.5, 2.5*sqrt(3)). Thus,
x^2 + y^2 = 9
(x-5.5)^2 + (y-2.5*sqrt(3))^2 = 25
where (x,y) are the coordinates of E. Solving we get
x = 69/98
y = 165/98*sqrt(3).
Now, if EAD = q then EHD = 2q where H is the center of circle alpha. Then
cos(2q) = x/3
cos(2q) = 2(cos(q))^2 - 1.
Solving gives cos(q) = 11/14.

• Typo 10 months ago

Unrelated comment: You look exactly like a youtuber I used to watch, he's called "YourMCAdmin"

• Aswini Banerjee 10 months ago

@Jeremy Weissmann it was just a mistake of my finger

• Jeremy Weissmann 10 months ago

Aswini Banerjee I don’t think you meant to post this as a reply.

• Hung Hin Sun 10 months ago

EFGD is a cyclic quadrilateral, angle FED=angle FGB=60 and angle EFG=angle EDA=90-angle EAD. Since angle DFG=30, so angle EFD=60-angle EAD. Also we can find DF=5sqrt(3) and ED=6sin(angle EAD). We can apply sine rule to triangle EDF and get (sin(60-angle EAD))/(6sin(angle EAD))=1/10. By using the compound angle formula and doing some simplifications, we can get tan(angle EAD)=5sqrt(3)/11. Hence we can find cos(angle EAD)=11/14.

• Minh Cong Nguyen 10 months ago

Let I be the midpoint of AD and let theta be the angle EAD, then angle EID is 2*theta.
In triangle IED we have cos(2*theta) =( 3^2+3^2-ED^2)/(2*3*3) = (18-ED^2)/18
In triangle IEG we have cos(2*theta) =( 3^2+8^2-EG^2)/(2*3*8) = (73-EG^2)/48
We have (18-ED^2)/18 = (73-EG^2)/48 (1)
Since EDGF is on the same circle, angle DEG = angle DFG,
GF=GD then angle DFG = angle FDG = 30°, therefore angle DEG = 30°
In triangle DEG we have cos(30°)=sqrt(3)/2 = (ED^2+EG^2-5^2)/(2*ED*EG) (2)
Solving for system of equations (1) and (2) where ED and EG are both positive, we have ED = 15*sqrt(3)/7 and EG=55/7, then cos(2*theta)=23/98, then cos(theta)^2=(1+cos(2*theta))/2 = 121/196, which means that cos(theta)=11/14

• Gabriel N. 10 months ago +2