Challenge: Can You Make All Three Angles Obtuse?

  • Published on Aug 16, 2018
  • Congratulations to Mohamed S. Fawzy, Benjamin Wang, Hung Hin Sun, Gabriel N., Allan Lago, GreenMeansGO, Phoenix Fire, Diego Viveros, Cesare Angeli, and Foster Sabatino for successfully solving the last week's math challenge question! Mohamed S. Fawzy was the first person to solve the question.
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    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.

Comments • 38

  • Bence Hervay
    Bence Hervay 10 months ago

    I got 9 (a=3, b=6)
    P can only be in the intersection of Thales-circles of segments AB, BC and CA and the area of this "spheringle" divided by the area of ABC is

  • Jean le Ronde d'Amelbert
    Jean le Ronde d'Amelbert 10 months ago

    I loved this weeks problem. You can easily show that if the point is within a Reuleaux triangle “inscribed” within the center of the triangle ABC then the point will satisfy the given condition. This is because of that one circle/right-triangle theorem. Finding the area of a Reuleaux triangle given its radius is trivial and finding the area of a unit equilateral triangle is also simple, so you can simply divide one by the other and simplify.

  • sachin singh
    sachin singh 10 months ago

    a+b= 9
    All the the angle should lie in a circle of radius (=1/2√3) inscribed in the main triangle ABC , now the probability is the area of circle divided by area of main triangle that is (π*(1/2√3)^2)/(0.5*1*√3/2) = √3π/9 ,which gives a=0 and b=9 .

  • Lim Chun Kiat
    Lim Chun Kiat 10 months ago


  • July_27 0_0
    July_27 0_0 10 months ago

    Can we use a calculator? ( ͡° ͜ʖ ͡°)

    • July_27 0_0
      July_27 0_0 10 months ago

      Ahh ok, Thx. However, because I am not really going in-depth yet in maths, I still stick to the basics, e.g. law of cosines, so yeah, thanks again for informing me 👍

    • LetsSolveMathProblems
      LetsSolveMathProblems  10 months ago

      Your solution should illustrate how to solve the problem without use of a calculator. However, I would probably accept solutions using programming languages for discrete math problems.

  • Joel Courtheyn
    Joel Courtheyn 10 months ago

    I also got 9 as the area wherein the point is to be found is bounded by circles with midpoint on the sides and radius equal to side length. This area is equal to z^2/16*(2*pi-2sqrt3). When we divide this by the area of the equilateral triangle : z^2 * sqrt3/4 we obtain : (pi*sqrt3 - 3)/6. The answer is thus 3+6 = 9.

  • Dwaraganathan Rengasamy
    Dwaraganathan Rengasamy 10 months ago

    Its 9.
    I drawn semicircles on each side and intersection of these inside the triangle will give us obtuse angle. Taking one side and centre of semicircle, connecting with the intersection of it with other sides, we get a kite. Finding its area and subtracting the sector area, we get the unwanted area. Now subtracting these from whole triangle and dividing by original triangle area, we get the required form giving us answer 9.

  • Wfreestyle
    Wfreestyle 10 months ago

    cool problem,too lazy to write the answer since many did haha

  • staffehn
    staffehn 10 months ago +2

    Answer is 9. I take a triangle with side lengths 2. A point P gives rise to an obtuse angle APB iff it lies inside the circle with diagonal AB (as we all know that the points on that circles border form right angles APB). We partition the triangle ABC into 4 smaller equilateral triangles with the midpoints of AB, BC and CA as the extra corners and side lengths 1. Then the circle around AB enters the triangle at BC’s midpoint, travels through the smaller triangle with corners C and the midpoints of BC and CA, and leaves ABC through the midpoint of CA; that is, it enters the smaller triangle with corner C through one of its corners and leaves through the other. The other 3 smaller triangles lie fully within that circle. Analogous observation for the other 2 circles determines what the area where P provides obtuse angles only looks like: it’s the intersection of the three circles with each other and ABC, including the center smaller triangle fully and a part of each outer smaller triangle. That extra part is exactly the area of a 60 degree circular segment with radius 1.
    Defining S = the area of a 60 deg circular sector with radius 1
    and T = the area of an side lenght 1 equilateral triangle,
    we have the full area of ABC being 4T, the area of the center small triangle T and the area of one of the 3 extra segments (S-T).
    That is, the area in question is:
    T + 3(S-T)
    and the probability is
    (T + 3(S-T)) / 4T.
    Now, T = sqrt(3)/4
    and S = 60/360 * π = π/6.
    So T + 3(S-T) = 3S-2T = 3π/6 - 2sqrt(3)/4 = π/2 - sqrt(3)/2.
    And 4T = sqrt(3).
    The probability then is:
    (T + 3(S-T)) / 4T = (π/2 - sqrt(3)/2)/sqrt(3) = π/2sqrt(3) - 1/2 = sqrt(3)π/6 - 1/2 = (sqrt(3)π - 3)/6
    And 3+6=9.

  • Quwertyn
    Quwertyn 10 months ago

    This probability is equal to the area of intersection of 3 circles with triangle sides as diameters divided by the area of the triangle. (On the sides of the circles one of the angles is π/2 and outside one of the angles is smaller so P has to be inside all of them). The intersection stays inside the triangle since if we step outside we get α+β

  • Assaf Abram
    Assaf Abram 10 months ago

    It’s cool to look at the inscribed releuax triangle

  • Evyatar Baranga
    Evyatar Baranga 10 months ago

    The solotion the overlapping area of three hafl circle siting on each line of the triangle ( because for each angle if it sits on the circle its 90 deg, and if its inside its bigger then 90)
    Tge circles go through the mid points of the triangle's lines, and so they divide it into triangels and roubded parts. All the raunded parts are the same: so in a half circle there are 3x+ 3 triangle/4. And the overlapping area is 3x + triangle/4
    So the overlapping are/ total are if the triangle = (sqrt (3)*pi -3)/6
    A = 3, b = 6
    Ans = 9

  • DynestiGTI
    DynestiGTI 10 months ago

    Damn. I thought it was a simple yes or no question.

  • Stephen Jiang
    Stephen Jiang 10 months ago +4

    do you write these questions yourself?

    • LetsSolveMathProblems
      LetsSolveMathProblems  10 months ago

      My pleasure! Thank you for your compliment. =)

    • Stephen Jiang
      Stephen Jiang 10 months ago +2

      LetsSolveMathProblems they’re very ingenuous, thanks for the videos

    • LetsSolveMathProblems
      LetsSolveMathProblems  10 months ago +3

      Most of the challenge problems (so far) were crafted by me. The rest are creation of my brother or both of us working in conjunction. I made this particular problem.

  • GreenMeansGO
    GreenMeansGO 10 months ago

    Yay. I got featured.

  • SloppyTaco
    SloppyTaco 10 months ago

    I got 9.
    Looking at the cases at which one of these angles is 90 degrees, the point has to lie on the circumference of a circle whose diameter consists of one of the edges of the equilateral triangle. The points that lie both on this circumference and within the triangle creates an arc that does from the midpoint of one edge to another. This logic applied to the other three creates a "triangle" whose edges are round. The area of this shape is (pi-sqrt(3))/8. To find the probability, divide by the total area to get (sqrt(3)*pi-3)/6. a+b is thus 3+6. Therefore, the answer is 9.

  • Abhinav Shripad
    Abhinav Shripad 10 months ago

    Ans9. Without losing generality assume the side length to be 1 if a point P satisfies given condition then it must lie inside the semicircle with diameter AB ,BC ,CA so the required probability is sum of semicircle area divided by total area which upon calculation gives
    So Ans is 9

  • Phoenix Fire
    Phoenix Fire 10 months ago

    First, I defined my triangle to have side length 2. I focused on 2 points and did a circle around the midpoint to see where the boundary of our shape is. Since the circle is of length 1, it's area is pi. Only a sixth of the circle is in our triangle, so we get pi/6. Subtract the unit equilateral to get pi/6-sqrt(3)/4. Add 2 of those to pi/6 to get (pi-sqrt(3))/2. Divide by the original triangle area of sqrt(3) aka multiply by sqrt(3)/3, and bring down the 3. Now we have (sqrt(3)pi-3)/6. 3+6=9

  • ehtuanK
    ehtuanK 10 months ago

    The Answer is 9:
    For an angle to be obtuse it must be within a semicircle over its corresponding side. So, for all angles to be obtuse P must be within the common arrea of the semicircles of all three sides. That is a reuleaux triangle (because our triangle is equilateral) and has an area of r²*(Pi - Sqrt(3))/2, where r is the radius of the semicircles. Since the triangle's sidelength is the semicircle's diameter, its area is (2r)²*Sqrt(3)/4 = r²*Sqrt(3). With P being chosen uniformly at random, its chance of being in the reuleaux triangle is the ratio of the two areas:
    (r²*(Pi/2 - Sqrt(3)/2)) / (r²*Sqrt(3)) = (Sqrt(3)*Pi - 3)/6
    So a=3 and b=6 and the answer is 9.

  • Marco Dalla Gasperina
    Marco Dalla Gasperina 10 months ago +2

    The point P has to fall within a region of 3 overlapping circles, centered at the midpoints of the sides. The area of the overlap looks like an inverted triangle with with bulging sides. The area of this bloated triangle (assuming a circle radius of 1) is 3pi/6 - 2*sqrt(3)/4 . (3 sector areas of angle pi/3 - 2 equilateral triangles of side 1). Divide this by the area of the big triangle which is sqrt 3 and the answer is (pi*sqrt(3) - 3)/6. Answer is 9.

  • Problematic(puzzle channel)

    Let the a be side of equilateral triangle. All the angles are obtuse when the point P falls into an area that is formed by intersecting of three circles each diameter being AB BC CA.
    This is because angle APB BPC CPA becomes circumference angle of circles with diameter being a. if P falls in the circle, then the angle becomes obtuse. If it falls out of the circle, then the angle becomes acute.
    The intersection shape is the equilateral triangle with side (a / 2) + 3 * (area of sector with angle 60 degrees and r = a / 2 - area of equilateral triangle with side (a / 2)). If you calculate this, it is √(3)/4 * (a/2)^2 + 3*(1/6 * (a/2)^2 * pi - √3/4 (a^2)). And we have to divide this by the area of the whole triangle, which is √3/4 * a^2. if you do this, the answer simplifies to( √3pi - 3) / 6. Therefore, a + b = 3 + 6= 9.

  • Albanovaphi7
    Albanovaphi7 10 months ago

    All angles are obtuse only if the P point is in a triangle of Reuleaux whose vertices are at the midpoints of the original triangle.

  • dominofan238
    dominofan238 10 months ago +6

    Let the length of one of the sides of the triangle be a. Draw three semicircles, each with radius a/2, so that their centers lie on the middle of the sides. Using Thales's theorem, we find that P must be located in the area A bounded by the three semicircles inside the triangle. This area consists of a smaller equilateral triangle of length a/2 and three lunes with area 1/6*π*(a/2)^2-sqrt(3)/4*(a/2)^2 each. We get A=a^2(π/8-sqrt(3)/8)) and, dividing by the area of the large triangle, the probability p=(sqrt(3)π-3)/6, so a=3 and b=6. The final answer is 3+6=9.

  • Allan Lago
    Allan Lago 10 months ago +1

    I got 9.
    Consider the semicircle with diameter BC. If P is at its circumference, BPC = 90, if P is outside the semicircle BPC 90.
    Generalizing this for any of the sides, for all three angles to be obtuse, P must be inside all of the semicircles with the sides of the triangle as their diameter.
    From there I used PIE to calculate the area of the intersection of the two semicircles as follows:
    Area of triangle = 3 * area of semicircle -3*intersection of two semicircles + intersection of three semicircles.
    For a side of length 2, Area of triangle = sqrt(3); area of semicircle = pi/2; intersection of two semicircles = 2pi/3 -sqrt(3)/2;
    Hence, intersection of three circles = (pi -sqrt(3))/2
    Divide by area of triangle to find the overall probability: P(all angles are obtuse) = (sqrt(3)*pi -3)/6
    a = 3, b=6. a+b = 9.

  • NoName
    NoName 10 months ago

    For an angle towards 2 selected points to be right, the 3rd, central point must be on a circle with diameter on selected 2 points.
    For angle to be obtuse, the 3rd point must be inside the circle
    Thus, we are searching for the relation between area of the shape between parts of semicircle and the area of the main triangle
    The shape we search is triangularly symmetric and all sides are parts of circle.
    Thus it consists of a equilateral triangle and 3 segments.
    If we set the side of main triangle as 2, the radius of circles would be 1. The area of main triangle is sqrt(3).
    Since the triangle with 2 vertexes same as equilateral and 1 being on the middle of its side is right one, the circles that we are interested in are going through centers of sides of main triangle
    Thus, the internal equilateral triangle has sides between middles of main triangle's sides.
    It's area is 4 times smaller that the main one's = sqrt(3)/4
    The segments are of the 60 degree angle (on equilateral triangle).
    Thus we can combine them with triangle to get segment.
    Combine them together and we get half of a circle. Area=pi/2
    To make the halfcircle we took 3 triangles - 2 are extra.
    So combined area of internal shape is pi/2-2*sqrt(3)/4=(pi-sqrt(3))/2
    And we need to divide it by the area of main triangle, sqrt(3)
    We get the chance of (pi-sqrt(3))/2sqrt(3)
    Or (sqrt(3)*pi-3)/6

  • Richard Chen
    Richard Chen 10 months ago

    Set the side length of ABC is 2, and set D, E, F are the mid-point of AB, BC, CA. Draw arcs DE, EF, FD based on the circle center on F, D, E. The probability will be AREA(DEF by arcs) / AREA(ABC). AREA(DEF by arcs) = AREA(Semicircle radius 1) - 2*AREA(Equaliteral Triangle side length 1)=(pi-sqrt(3))/2, AREA(ABC)=sqrt(3). So the Probability is (pi-sqrt(3))/(2*sqrt(3))=(sqrt(3)*pi-3)/6.

  • Beshoy Nabil
    Beshoy Nabil 10 months ago

    Ans =9
    Suppose the side length of the equilateral triangle is (2x) and D,E & F are the midpoints of AB,BC & AC respectively. Draw a circular arc of diameter AB, center D, angle 60 deg and intersects the triangle at E & F. Angle APB will be obtuse if and only if P is inside the circle. Repeat the process and draw arcs DF & DE of centers E & F respectively. The bounded area by the three arcs are the required region of P and the required probability is this area over the area of equilateral triangle ABC.
    Bounded area DEF = 3*(sector of angle 60 - equilateral triangle DEF) + equilateral triangle DEF
    = 3* ((pi*x^2*60/360) - (0.5*x^2*sin(60)) + root(3)*x^2/4 = (pi-root(3))*x^2/2
    Area of ABC = 0.5*(2x)^2*root(3)/2=root(3)*x^2
    Probability = ((pi-root(3))*x^2/2) / (root(3)*x^2) = (root(3)*pi - 3)/6
    then a=3, b=6 & a+b=9

  • Throxs
    Throxs 10 months ago

    you can draw three different arcs which will give you a right-angle triangle... If you pick a point inside these three arcs, the resulting triangles will all be obtuse... To calculate the desired probability we just devide the area between the arcs by the total area of the triangles... This results in the formula P = (√3π - 3)/6... Therefore a + b = 9

  • attyfarbuckle
    attyfarbuckle 10 months ago

    3 + 6 = 9. The shape is the union of 3 sixth-circles. Delete two triangles for the overlap.

  • Zain Majumder
    Zain Majumder 10 months ago

    Construct a circle on one side of the triangle (AB, for example) whose diameter is equal to AB and whose center is the midpoint of AB. By inscribed angles, any point P selected on this circle will form a 90 degree angle APB, so any point P selected within the circle will have obtuse angle APB. The same logic applies to the other side, so the probability is the fraction of the triangle which is in all three circles. This area looks like a round triangle (3 60 degree sectors overlapping). Answer is (3(1/6)(pi*r^2)-2(r^2sqrt3/4))/((2r)^2sqrt3/4) = (pi*sqrt3-3)/6. Answer is 9.

  • diego perez salinas
    diego perez salinas 10 months ago

    If we look the problem angle by angle , we realize that the sector of the points that satisface the conditions is an equilateral triangkw (each side equal half the side of ABC)with 3 circular sector of 60 degrees in each side of the triangkw . Calculating this area and diving by the area of ABC gives (pi*sqr(3)-3)/6 . We conclude that a+b=9

  • Benjamin Wang
    Benjamin Wang 10 months ago

    Answer is 9. Let AB=2. Let midpoints if AB, BC, CA be D, E, F respectively. Draw semicircles centres at D, E, F. The area of the rounded triangle shape S in the middle (vertices DEF, sides are arcs of circles) is (pi-sqrt3)/2. The area of the whole triangle ABC is sqrt3. Since angles on a semicircle = right angle, any P inside S will satisfy the constraints. The ratio of areas imply that a=3, b=6.

  • Kevin M
    Kevin M 10 months ago +1

    The probability of APB being obtuse is the probability of P being in the circle with diameter AB, and similarly for BPC and APC. Let the triangle have side length 2 for simplicity. The probability P is in all 3 circles with diameter AB, BC, AC is 3*(1^2*π/6)-2*((√3)/2), divided by the area of the triangle. This equals (π-(√3))/2/(√3)=(π(√3)-3)/6. A+b is 9