3 Circles and a Line

Geometry Level 2

Three unit circles are drawn tangent to each other with all their centers lying on line segment O E . OE.

From O , O, line segment O D OD is drawn such that it's tangent to the rightmost circle.

If the length of the chord A B AB is a b \frac ab , where a a and b b are coprime positive integers, then what is a + b ? a+b?


Note: I did not create this problem; I simply solved it and decided to post it on here. Credit goes to my math teacher.


The answer is 13.

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12 solutions

Abraham Zhang
Dec 2, 2018

sin ( α ) = D F O F = 1 5 O G sin ( β ) = A G sin ( α ) sin ( β ) = 3 5 G B 2 = A B 2 + A G 2 2 A B × A G cos ( β ) A B = 8 5 \begin{aligned} \sin(\alpha) &= \frac{DF}{OF}\\ &=\frac15 \\ \frac{OG}{\sin(\beta)}&=\frac{AG}{\sin(\alpha)} \\ \sin(\beta)&=\frac35 \\ GB^2 &= AB^2+AG^2-2AB\times AG\cos(\beta) \\ AB &= \frac85 \end{aligned}

Why does sin of alpha equal DF over OF?

Adrian Self - 2 years, 6 months ago

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In a right angle triangle, sine = opposite hypotenuse =\frac{\text{opposite}}{\text{hypotenuse}} .

Abraham Zhang - 2 years, 6 months ago

Tangents are perpendicular to the radius sharing the common point

Rac Arora - 2 years, 6 months ago

How do I get the value of cos beta? Or is it always 4/5 when sin is 3/5?

Rachel Tam - 2 years, 6 months ago

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If sin ( β ) = 3 5 \sin (\beta)=\frac35 , then cos ( β ) = 4 5 \cos (\beta)=\frac45 when β \beta is in the first quadrant.

Abraham Zhang - 2 years, 6 months ago

You can get the value of cos(b) if you draw a right triangle with ratio from sin(b), and use the Pythagorean theorem to calculate the length of the leftover side. Also, you can calculate the value from the formula sin^2(a)+cos^2(a)=1. This formula allows you to calculate the value of sin or cosine when you have the value of the other one(with the same angle).

Nether Deth - 1 year, 9 months ago

Can someone explain the second line (OG/sin(beta)=AG/sin(alpha))?

Orrin Ahola - 2 years, 6 months ago

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sin(pi-beta)=sin(beta)

chen aoton - 2 years, 6 months ago

It is the sine rule.

Abraham Zhang - 2 years, 6 months ago

This is awesome ❤️ I did it in tedious manner with Cartesian geometry

Karthik Rambhatla - 2 years, 6 months ago

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I think coordinate geometry way is way cooler. Far more maths involved and more learning for my students than the trig way. Many of my students could do the trig way but they all learned something the coordinate geometry way. Give yourself some credit! :-)

Simon Dixon - 1 year, 12 months ago

Why OG/sin(β) = AG/sin(α) ?? Someone said it's the law of sines, but these segments aren't opposite to the referenced angles; and someone else talked about right triangles, and these aren't right triangles. I'm still confused by that line.

Omar Balbuena - 2 years, 6 months ago

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The angle opposite to O G OG is equal to π β \pi-\beta . In my use of the sine rule, the sin ( β ) = sin ( π β ) \sin(\beta)=\sin(\pi-\beta) .

Angle D D is a right angle since O D OD is tangent to the circle that passes through D D .

Abraham Zhang - 2 years, 6 months ago

This has a nice solution by applying inversion with respect to central circle. No trigonometry needed in that case.

Jonas Pauliukevicius - 2 years, 5 months ago

Abraham really should have mentioned 2 points: Angle ODF is 90 degrees so sin(ODF) = 1 Triangle ABG is isosceles, and he really should have placed beta at angle ABG to use the sine rule. These points are rather obvious though.

Kimberly Rae - 2 years, 5 months ago

GB^2 = AB^2 + AG^2-2 AB AG cos(β) ?

Kaustav Bhattacharya - 2 years, 2 months ago

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Cosine Rule

Abraham Zhang - 2 years, 2 months ago
Ajit Athle
Nov 23, 2018

Let l=AB. Draw a perpendicular GH from G to AB. Note that H is the midpoint of AB since GH is drawn perpendicular to AB. Moreover, triangles OGH & OFD are similar hence: G H F D \frac{GH}{FD} = 3 5 \frac{3}{5} ; hence GH = 3 5 \frac{3}{5} since FD =1. Now,from right tr. AGH, l 2 4 \frac{l^2}{4} =1 - 3 2 5 2 \frac{3^2}{5^2} or l = AB= 8 5 \frac{8}{5} .

Yes! This is the solution I used! You might need to include a few details, though.

Blan Morrison - 2 years, 6 months ago

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It's great for the 3rd grade trigonometry lesson, so I'll include it in my notebook (if you don't mind). I like Zhang's solution because he used both sine and cosine rules and a useful property sin x = sin (180 - x) (and, of course, the right triangle definition of the sine). Thank you very much for sharing this problem.

lovro cupic - 2 years, 6 months ago

This is by far the simplest method to solve this problem. You might need to add a few details so other people can follow your logic. The similar triangles are OHG and ODF. The last line refers to another right angled triangle AHG. Since AH is half of AB, the answer is AH x2.

Pleayo Tovaranonte - 2 years, 6 months ago

"Note that H is the midpoint of AB since" AG and BG are equal, (so AGB is an isosceles triangle) and "GH is drawn perpendicular to AB"

Robert Andrzejuk - 2 years, 6 months ago

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Yes, and...?

Blan Morrison - 2 years, 6 months ago

It's how I did it, but I made a mistake in the very last step, I forgot to take the square root basically. I calculated AH (but ACTUALLY calculated AH², which is 16/25), then took twice that to get a length AB of 32/25. oops

Iskander Elderson - 2 years, 6 months ago

Lets plot this figure on Cartesian Plane. O O will be Origin.

We get G = ( 3 , 0 ) G = (3,0) and F = ( 5 , 0 ) F = (5,0)

Equation of circle with center G G will be ( x 3 ) 2 + y 2 = 1 (x - 3)^2 + y^2 = 1

Equation of Circle with center F F will be ( x 5 ) 2 + y 2 = 1 (x - 5)^2 + y^2 = 1

Let the tangent line meet the Circle ( F F centered) at D ( p , q ) D(p,q)

In traingle O D F ODF , using Phythagoras Theorem, we get O D = 5 2 1 2 = 2 6 OD = \sqrt {5^2 - 1^2} = 2\sqrt6

To calculate p p and q q , we have folllowing equations,

p 2 + q 2 = 24 p^2 + q^2 = 24 and ( p 5 ) 2 + q 2 = 1 (p - 5)^2 + q^2 = 1

We get q = 24 5 q = \dfrac {\sqrt{24}}{5} and p = 24 5 p = \dfrac {24}{5}

We get the equation of the tangent line as x = 24 y x = \sqrt{24} y

We now find the points of intersections of this line with ( x 3 ) 2 + y 2 = 1 (x - 3)^2 + y^2 = 1

We get two points as B = ( 24 ( 3 24 + 4 25 ) , 3 24 + 4 25 ) B = (\sqrt{24}(\dfrac {3\sqrt{24} + 4}{25}) , \dfrac{3\sqrt{24} + 4}{25}) and A = ( 24 ( 3 24 4 25 ) , 3 24 4 25 ) A =(\sqrt{24}(\dfrac {3\sqrt{24} - 4}{25}) , \dfrac {3\sqrt{24} - 4}{25})

We get A B = 8 25 ( 24 ) 2 + 1 2 = 8 25 × 5 = 8 5 = 8 5 = a b AB = \dfrac {8}{25} \sqrt{(\sqrt{24})^2 + 1^2} = \dfrac {8}{25}\times 5 = \dfrac {8}{5} = \boxed{\dfrac {8}{5} = \dfrac {a}{b} }

Therefore, a + b = 8 + 5 = 13 a + b = 8 + 5 = \boxed{13}

Interesting take with the Cartesian plane! Never would've thought of that approach!

Blan Morrison - 2 years, 6 months ago

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You can even solve 3-D objects having symmetry using this. Now, you can even calculate AF, BF ,etc if you want!

A Former Brilliant Member - 2 years, 6 months ago

Where did you get any numbers from in the first place? I'm sure its logical it's just not clicking for me

wes mok - 2 years, 6 months ago

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Sorry, I didn't get which numbers you are talking about. Can you be more precise please?

A Former Brilliant Member - 2 years, 6 months ago

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The description doesn't give any number for me to work with however denis says the radius is 1. Where did the 1 come from?

wes mok - 2 years, 6 months ago

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@Wes Mok Pick any other number and you'll end up simplifying a the and and get an answer equal to the answer you would've gotten if you picked radius = 1 from the beginning.

denis vladu - 2 years, 6 months ago

@Wes Mok The question states three "unit" circles. Here, "unit " implied that they have radius equals unity which is definitely equals 1.

A Former Brilliant Member - 2 years, 6 months ago

Radius = 1.

denis vladu - 2 years, 6 months ago

I did everything same except the OD part. I misinterpreted as hypotenuse and ended up with some radicals😁

Karthik Rambhatla - 2 years, 6 months ago
Binky Mh
Dec 3, 2018

By comparing to similar triangle O D F \triangle ODF , we can see that the line from G G which bisects the chord A B AB has a length of 3 5 \frac{3}{5} . Using Pythag, we can work out the length of A B 2 \frac{AB}{2} :

( A B 2 ) 2 = 1 2 0. 6 2 ( A B 2 ) 2 = 1 0.36 ( A B 2 ) 2 = 0.64 A B 2 = 0.8 A B = 1.6 = 8 5 8 + 5 = 13 \begin{aligned}(\frac{AB}{2})^2&=1^2 - 0.6^2\\ (\frac{AB}{2})^2&=1 - 0.36\\ (\frac{AB}{2})^2&=0.64\\ \frac{AB}{2} &= 0.8\\ AB =& 1.6 = \frac{8}{5}\\ 8+5 & =\boxed{13}\end{aligned}

This is exactly how I solved it!

Blan Morrison - 2 years, 6 months ago
Dhruva Ahuja
Dec 4, 2018

OHG ~ ODF (AA similarity)

=> O G O F \frac{OG}{OF} = G H F D \frac{GH}{FD}

We know OG = 3 units, OF = 5 units and FD = 1 unit => HG = 3 5 \frac{3}{5} units

If the perpendicular distance of a chord from the centre is given then length of the chord can be calculated easily using Pythagoras theorem.

=> AB= 2*sqrt(1- 9 25 \frac{9}{25} ) = 8 5 \frac{8}{5} .

nice answers!

Sam Wang - 1 year, 9 months ago
David Vreken
Nov 22, 2018

From trigonometry on D E F \triangle DEF we have O D = 2 6 OD = 2\sqrt{6} and cos O = 2 6 5 \cos O = \frac{2\sqrt{6}}{5} .

From the power of a point theorem , the two secants to circle G G have the relation O A O B = 2 4 OA \cdot OB = 2 \cdot 4 , or O B = 8 O A OB = \frac{8}{OA} .

From the law of cosines on O B G \triangle OBG , B G 2 = O B 2 + O G 2 2 O G O B cos O BG^2 = OB^2 + OG^2 - 2 \cdot OG \cdot OB \cdot \cos O , or 1 2 = ( 8 O A ) 2 + 3 2 2 3 8 O A 2 6 5 1^2 = (\frac{8}{OA})^2 + 3^2 - 2 \cdot 3 \cdot \frac{8}{OA} \cdot \frac{2\sqrt{6}}{5} , which solves to O A = 6 6 ± 4 5 OA = \frac{6\sqrt{6} \pm 4}{5} .

This makes O B = 8 O A = 6 6 4 5 OB = \frac{8}{OA} = \frac{6\sqrt{6} \mp 4}{5} , which means A B = O B O A = 8 5 AB = OB - OA = \frac{8}{5} . Therefore, a = 8 a = 8 and b = 5 b = 5 , and a + b = 13 a + b = \boxed{13} .

Draw a perpendicular GH from G to AB and let l=AB. Now by similar triangles: GH/FD =3/5; hence GH =3/5 since FD =1. Now, l²/4 =1 - (3/5)² or l = AB= 8/5.

Ajit Athle - 2 years, 6 months ago

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Great solution! You should post it in the solutions section so people can upvote it.

David Vreken - 2 years, 6 months ago

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Seconded - much neater than my solution! (As I mention in another comment, it's really interesting to see so many different approaches)

Chris Lewis - 2 years, 6 months ago

Okay thanks, guys. I've posted it as advised.

Ajit Athle - 2 years, 6 months ago
Rogerio De Souza
Dec 4, 2018

Find below my solution

Chris Lewis
Nov 23, 2018

If you were told, "find a point P P on the line O D OD satisfying P G = 1 PG=1 ," you would spot that there are two such points: A A and B B . This ambiguity is an example of the solution of a triangle Δ P O G \Delta POG , knowing only two sides and a non-included angle (see here , for example). We have the following information about Δ P O G \Delta POG :

  • O G = 3 OG=3 (since the circles have unit radius)

  • P G = 1 PG=1

  • sin P O G = 1 5 \sin{\angle POG}=\frac{1}{5} (since P O G = D O F \angle POG=\angle DOF , O F = 5 OF=5 , D F = 1 DF=1 and O D F = 9 0 \angle ODF=90^{\circ} )

We're interested in the side length O P OP ; let's call this x x . By the cosine rule, P G 2 = O P 2 + O G 2 2 O P O G cos P O G PG^2 = OP^2+OG^2-2OP \cdot OG \cos{\angle POG} .

Substituting, this is 1 2 = x 2 + 3 2 6 x cos P O G 1^2=x^2+3^2-6x\cos{\angle POG} . Note that this is a quadratic in x x , with - as we'd expect - two solutions.

Rearranging and completing the square, this becomes x 2 6 x cos P O G + 9 cos 2 P O G = 8 + 9 cos 2 P O G x^2-6x \cos{\angle POG} + 9\cos^2 {\angle POG} = -8+9\cos^2 {\angle POG}

Plugging in sin P O G = 1 5 \sin{\angle POG}=\frac{1}{5} this becomes ( x 3 cos P O G ) 2 = 8 + 9 ( 1 1 25 ) = 16 25 (x-3 \cos{\angle POG})^2 = -8+9(1-\frac{1}{25}) = \frac{16}{25}

Taking square roots, x = 3 cos P O G ± 4 5 x=3 \cos{\angle POG} \pm \frac{4}{5}

The two solutions correspond to the lengths O A OA and O B OB ; we want A B AB , the difference of these, which is just

( 3 cos P O G + 4 5 ) ( 3 cos P O G 4 5 ) = 8 5 (3 \cos{\angle POG} + \frac{4}{5})-(3 \cos{\angle POG} - \frac{4}{5}) = \frac{8}{5} , so the required answer is 8 + 5 = 13 8+5=\boxed{13}

Nice solution! It was very similar to mine, but the only trigonometry I used was the Pythagorean Theorem.

Blan Morrison - 2 years, 6 months ago

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Thanks - you should post your solution! I find it really interesting to see different approaches. I originally used coordinate geometry, as per Niraj Sawant's solution - after all, you called one of the points O O , so how could I resist? - but then started trying to angle hunt A G B \angle AGB in order to find the chord length; that led to my posted solution. Fun fact: the length of the chord in the leftmost circle is irrational. I wonder what makes the middle circle special. If we had different numbers of circles, would we still find rational chords?

Chris Lewis - 2 years, 6 months ago
Ralph Gatan
Dec 9, 2018

The solution lies on our perspective as O is 1 and A is 2 B is 3 but actually I just a pythagorean theorem thingy 2^2 + 3^2 = 13 or c^2

I think you missed a few steps. Also, that is just incorrect.

Blan Morrison - 2 years, 6 months ago
Alex Burgess
Dec 7, 2018

Let X be the midpoint of AB. GX is perpendicular to AB.

GOX is similar to FOD, so GX=FD*GO/FO=3/5.

AX²=AG²-GX²=1²-(3/5)²=(4/5)².

AB=2AB=8/5.

Vinod Kumar
Dec 4, 2018

Drop a perpendicular from G on chord AB, which can be easily proved=(3/5). Use Pythagoras to show

AB=2(√1-(9/25))=(8/5)

Answer=13

Edwin Gray
Dec 3, 2018

Let t be the mid-point of AB. Tan(<DOF) = 1/2.5 = .4. TG/OG = .4 = TG/1.5, TG = .6, TB = .8, AB = 1.6 = 8/5, 8 + 5 = 13. Ed Gray

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