A Challenging Geometry Problem from Russia

Рет қаралды 2,190

Math Window

2 ай бұрын

Are you able to solve this geometry challenge? Watch the video and learn how to solve!

Пікірлер: 13

@romank.6813 2 ай бұрын

So, we've got a right triangle with the height 1 and the hypotenuse 4. That means that the radius of the circumscribing circle is 2. That means that the smaller cathetus of the the triangle subtends an arc of 30° (since 1/2=sin(30°)). The angle under question is an inscribed angle subtending the same arc. Therefore, it is 15°.

@franolich3 2 ай бұрын

A simpler solution: Let the required angle be x. The shortest triangle length = 4.sin(x) The angle between the shortest triangle length and the rectangle vertical is also x. So (4.sin(x)).cos(x) = 1. => sin(2x) = 1/2 => x = 15

@blatetastic 2 ай бұрын

just use angle at centre twice angle at circumference after you found theta

@Braxton-oz5sr 2 ай бұрын

Couldn’t you just use tangent after creating the first right triangle?

@shadrana1 2 ай бұрын

Compare small and large upper right triangles which are similar. x/1=1/(4-x) using the tangents. 4x-x^2=1 x^2-4x+1=0..........................................(1) (x-(2+sqrt3)(x-(2-sqrt3))=0 x= (2+sqrt3) or x= (2-sqrt3) (2+sqrt3)>1 hence this is ruled out. x= (2-sqrt3) check out equation (1) for x=(2-sqrt3) x^2= 7-4sqrt3 -4x= -8+4sqrt3 1=1 x^2-4x+1= 7-4sqrt3 -8+4sqrt3+1=0 as required tan (?)=1/(4-x)=1/(4-(2-sqrt3)) = 1/(2+sqrt3) = arctan(2-sqrt3)=15 deg. ( standard exact value of tan15deg.=(2-sqrt3)) (?)=arctan(1/(2+sqrt3))=15 deg. and that is our answer. Thanks for the puzzle Math Window.

@prime423 2 ай бұрын

A problem like this requires some insight!!Romank is correct-an easy solution. For most, we are asked for an angle. Look at the given info. A side 1 with a radius 2.A 30-60-90 right triangle. The rest is trivial.

@allanflippin2453 2 ай бұрын

Ms Math Window, Your solution is all clear except the step where you decide the length of the pink line is 2. Of course it is correct, but I can't follow how to prove is it 2?

@derwolf7810 2 ай бұрын

It is Thales's theorem. The vertex with the right angle must be located on the (half) circle above the hypotenuse and the midpoint of the hypotenuse is the center of that circle.

@allanflippin2453 2 ай бұрын

@@derwolf7810 Got it, thanks!

@thichhochoi766 2 ай бұрын

You are dead wrong at 3:24. The triangle is isosceles because the two side are qual, not because the 2 angles are qual. It is totally different.

@adipy8912 Ай бұрын

You are true when saying that it is isosceles if 2 sides are equal. But if 2 sides are equal, then 2 angles will automatically be equal. Therefore saying that it is not isosceles because of 2 equal angles doesn't make sense. He is not dead wrong.

@thichhochoi766 Ай бұрын

@@adipy8912 Ha ha ha. Listen carefully. She said, "the triangle are isosceles because the 2 angles are equal", while she does not know that those 2 angles are equal. That is the dead wrong part.