Solving a viral geometry problem from Twitter

Are you able to resolve for the world of the sq. and the size of the tangent line? Particular thanks this month to Fiona Harper, Kyle, Mike Robertson, Michael Anvari, Robert Zarnke. Due to all supporters on Patreon!

Tailored from a puzzle by Twitter @Cshearer41 (Catriona Agg)

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Solving a viral geometry problem from Twitter

38 thoughts on “Solving a viral geometry problem from Twitter”

  1. I think the solution is incomplete. Hence, you haven't provide a proof that the two x's are equal. In my point of view the sides of the triangle are 12+x and 6+y. You are still requested to show whether x=y or not.

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  2. I felt great solving both parts of the problem after five hours or so. For me, part 2 (finding AB) is pretty quick though. It's a great geometry contest problem!

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  3. For the first part I did 2x² + 36x – 144 = 0
    Halve it -> x² + 18x – 72 = 0
    Complete a² + 2ab + b² -> x² + 2*9*x + 9² – 81 – 72 = 0
    So it turns into (x + 9)² – 153 = 0
    Move independent part to right and sqrt (x + 9) = ± √153
    x + 9 = ±√9*17 = ±3√17
    x = -9 ± 3√17
    Only use positive because we're measuring distance.
    x = 3√17 – 9

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  4. Finally finished one of these questions and got it right! I did it using a lot of algebra using the change in y and x from the centre of the smaller circle to the intersection of the circles, and called them n and m respectively, then found the gradient of the line going downwards, found the negative reciprocal, then used that to find the 2 points on the line, then used pythagoras' theorem to find the distance, my method was much harder but I'm glad I actually came to an answer, a correct one at that, for one of these questions

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  5. I solved this using algebra but the calculation turns out to be kinda messy and complicated. Even so, I'd like to share my method here:
    Take that corner of the square which lies on the semi circle to be O(0,0), the corner nearest to the semi circle which does not lie on the semicircle to be P [This is the corner on which the right triangle was made in the video] and the corner nearest to the quarter circle but not lying on it (other than P) to be Q. The positive x direction is OP and positive y direction is OQ. So the center of the small circle is at A(6,0).

    Let us assume there is a point T(a,b) on the small circle and there is another point B collinear to AT such that T divides AB in the ratio 1:2. [The large quarter circle in the diagram is centered at B with BT as the radius and T being the tangent point between the circles, hence the specific ratio of division]. Now we can find the value of B in terms of a, b as B(3a-12,3b).

    Now P lies on x axis hence its y coordinate is 0, and as OP is perpendicular to BP, x coordinate of P and B is equal. Thus P is (3a-12, 0). So, the length of the square side is 3a-12=3b => b=a-4. Thus T is expressed as (a,a-4).

    Now we know length of AT = 6, A(6,0), T(a,a-4). We solve for a using distance fomula and quadratic equation and get a=5±√17. Since a-4>0 (due to the way x and y axes are defined), a=5+√17.

    From this, side length of the square = 3a-12 = 3√17 + 3. Thus the area = (3√17 + 3)² = 162 + 18√17

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  6. Geometry and Trigonometry… I saw the answer for the length of AB, but just couldn't figure HTF I got there lolol. The area of the square was easy using basic Geometry.

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  7. Since AB is a tangent line and it is on the parallel edge of the square so that it's the same as the line between two center of circle. that means it's 18.

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  8. You can also solve it by seeing that lengths of semicircle and quarter-circle are the same, from there you will see that what you call alpha-angle is equal to 30deg, and that is all you need to know.

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  9. In a square, all angles are 90° and all sides are of equal length. Therefore, any perpendicular line segments connecting opposite sides have the same lengths.

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  10. For the 2nd part, I spent a lengthy series of calcs with many decimals. And I got it wrong.
    I can't believe how easily and cleanly you did it! Very nice!

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  11. There is a quicker way: Let C and D be the two centers of circles. B' the corner forming a right-angled triangle with C and D, A' the point such that ABB'A' is a parallelogram and E corner point on the smaller circle. We have that AB is perpendicular to CD, but then so is A'B', thus <EB'A'=<B'DC and hence triangle EB'A' is congruent to B'DC. Thus, AB=A'B'=CD=18.

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  12. all parallel lines that touch the same opposite sides of a square are of the same length.
    we know the line between the centre of 2 circles equals 18 (call it CD)and it is perpendicular to AB thus these 2 lines also form the same angle with the sides they intersect. let's rotate AB 90degrees and it becomes a line parallel to CD and touches the same sides of the square hence AB = 18.

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  13. That's really doing it the hard way. Let P be the point where the circles touch. Draw a horizontal line through P. The angle between this line and AB equals the angle that the line through the centres and P makes with the right side of the square. Call this angle beta. Let d be the length of the side of the square, then cos(beta) = d/18. But AB = d/cos(beta) = 18.

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  14. Actually I could find AB straight away because it's perpendicular to the segment connecting the centres.
    Solving for the area is more time consuming because of the tricky quadratic equation.

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  15. Finally, FINALLY managed one (half) of these problems – part 2, by just looking at the thumbnail. I somehow got AB was the same length as those radiuses, being at right angle to them and ending at opposite edges of the square. That's all 😛 (maybe I was just lucky).

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