Challenge: Can You Make All Three Angles Obtuse?

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.

[jeanlerondedamelbert]

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.

[assafabram]

It’s cool to look at the inscribed releuax triangle

[SlipperyTeeth]

9
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

[finweman]

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.

[JoelCourtheyn]

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.

[lucas]

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.

[staffehn]

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.

[kevinm]

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

[evyatarbaranga]

9

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

[ehtuanK]

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.

[dwaraganathanrengasamy]

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.

[sachinsingh-wbdn]

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 .

[Quwertyn]

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<π so α<π/2 or β<π/2. It also touches each side in it's middle since at that point one angle is π and two are π/2. We can easily calculate the area of the "cut off" part of the big triangle. It's equal to big triangle-two small triangles-one sixth of a circle so:
w=half of the side of the triangle

The area of the intersection=big triangle-3 cutoffs so
w^2sqrt(3)-3w^2/2(sqrt (3)-π/3)=w^2/2(π-sqrt (3))
We just have to divide it by the area of the big triangle now.
w^2/2(π-sqrt(3))/w^2sqrt(3)=
(π-sqrt(3))/2sqrt(3)=
(πsqrt(3)-3)/6
Therefore a=3 and b=6. a+b=9.

[allanlago]

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, and if P is inside of 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.

[bencehervay]

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
(3•(π/6-√3/4)+√3/4)/√3=
=π/(2√3)-1/2=
=√3•(pi-√3)/(2•3)=
=(π√3-3)/6

[aldolunabueno]

a+b=9
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.

[randomchannel]

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.

[DynestiGTI]

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

[problematicpuzzlechannel]

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.

[AQWraghd]

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