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geometry question: Given a triangle - all sides unequal. Place a semicircle entirely inside the triangle. What is maximum possible semicircle for prescribed triangle? Describe semicircle by locations of end points and midpoint.

c--------

The maximum possible semicircle that can be placed inside a triangle with all sides unequal is known as the incenter circle and is tangent to all three sides of the triangle. The endpoints of the diameter of the semicircle lie on the two sides of the triangle that form the smallest angle and the midpoint of the diameter is the incenter of the triangle, which is the point where the angle bisectors of the three angles of the triangle meet.

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geometry question: Given a triangle - all sides unequal. Place a semicircle entirely inside the triangle. What is maximum possible semicircle for prescribed triangle? Describe semicircle by locations of end points and midpoint.

c------------

To maximize the size of the semicircle, we want to place it such that its diameter is as long as possible. Since the semicircle must be entirely inside the triangle, we can draw the three perpendicular bisectors of the sides of the triangle. These lines will intersect inside the triangle at a single point, called the circumcenter.

The circumcenter is equidistant from the three vertices of the triangle, so the diameter of the maximum semicircle will be the distance from the circumcenter to the farthest vertex.

Therefore, the endpoints of the semicircle will be where the diameter intersects the two sides of the triangle adjacent to the farthest vertex. The midpoint of the semicircle will be the circumcenter.

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When one of the angles is obtuse, the circumcenter will be outside the triangle - semicircle partially outside.

c-----------------------

This is because the circumcenter is the point where the perpendicular bisectors of the sides of the triangle intersect, and when one angle is obtuse, the perpendicular bisector of the side opposite that angle will intersect outside the triangle. This results in a semicircle that extends partially outside the triangle.

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Your original answer is incorrect, since semicircle must be entirely inside triangle.

c--------------------

I apologize for the mistake in my original answer. You are correct that a semicircle inscribed in a triangle must be entirely inside the triangle. This means that the diameter of the semicircle cannot be longer than the shortest side of the triangle and its center must also be inside the triangle.

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Correct answer? Place diameter on longest side withe arc tangent to other two sides.

c------------------------------------

This answer is incomplete and cannot be determined without additional context. It appears to be related to a specific geometry problem or construction. Please provide more information for a complete answer.

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For given sides, the angles are known and the center and radius can be gotten from the angles and the length of the longest side.

c----------------------------

This statement is not entirely accurate. In a triangle, the center and radius of the circumcircle (the circle that passes through all three vertices of the triangle) can be determined from the lengths of the sides or the angles of the triangle, but not necessarily both at the same time.

To find the center and radius of the circumcircle using the angles of the triangle, you can use the following formulae:

The center of the circumcircle is given by the intersection of the perpendicular bisectors of any two sides of the triangle. The perpendicular bisector of a side is the line that passes through the midpoint of the side and is perpendicular to it.

The radius of the circumcircle can be found using the formula:

r = a/(2sinA)

where a is the length of the longest side of the triangle, and A is the angle opposite to side a.

To find the center and radius of the circumcircle using the lengths of the sides of the triangle, you can use the following formulae:

The center of the circumcircle is given by the point of intersection of the three perpendicular bisectors of the sides of the triangle.

The radius of the circumcircle can be found using the formula:

r = abc/(4Δ)

where a, b, and c are the lengths of the sides of the triangle, and Δ is the area of the triangle, which can be found using Heron's formula.

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Wrong answer! Question has semicircle entirely inside triangle. Your answer has it partially outside.

c-------------------

I'm sorry for the mistake in my previous answer. Here's an updated solution:

Let the radius of the semicircle be r, and let the midpoint of the base of the triangle be O. Draw the perpendicular from O to the semicircle, and call the point where it intersects the semicircle A. From A, draw two lines tangent to the semicircle, one to each endpoint of the base of the triangle. Label the points where these lines intersect the base of the triangle B and C.

Since OA is perpendicular to AB and AC, then triangle OAB and triangle OAC are right triangles. Therefore, we have:

OA^2 + AB^2 = OB^2

OA^2 + AC^2 = OC^2

Subtracting these equations, we get:

AB^2 - AC^2 = OB^2 - OC^2

But we know that OB = OC (since they are both radii of the semicircle), so this simplifies to:

AB^2 - AC^2 = 0

Thus, AB = AC. Let this common length be x. Then we also have:

AO = AB/2 = AC/2 = x/2

Let the height of the triangle be h. Then we have:

r = AO - h

r = x/2 - h

Equating these two expressions for r, we get:

x/2 - h = AO - h

x/2 = AO

x = 2AO

But we know that AO = h/(2tan(theta)), where theta is half the angle at the vertex of the triangle opposite the base. Therefore:

x = h/tan(theta)

So the area of the triangle is:

A = (1/2)bh

A = (1/2)(x)(h)

A = (1/2)(h/tan(theta))(h)

A = (1/2)h^2/tan(theta)

And the area of the semicircle is:

B = (1/2)pir^2

B = (1/2)pi(x/2-h)^2

B = (1/8)pi*(x^2 - 4hx + 4h^2)

Thus, the ratio of the area of the semicircle to the area of the triangle is:

B/A = [(1/8)pi*(x^2 - 4hx + 4h^2)]/[(1/2)h^2/tan(theta)]

B/A = (pi/4)[(x^2 - 4hx + 4h^2)/(h^2/tan(theta))]

B/A = (pi/4)[2tan(theta)(x^2 - 4hx + 4h^2)/h^2]

B/A = (pi/2)[(x^2 - 4hx + 4h^2)/(h^2tan(theta))]

B/A = (pi/2)[x^2/h^2 - 4x/h + 4]

Substituting x = 2h/tan(theta), we get:

B/A = (pi/2)[(4/tan^2(theta)) - (8/tan(theta)) + 4]

B/A = (pi/2)[(4sin^2(theta)/cos^2(theta)) - (8sin(theta)/cos(theta)) + 4]

B/A = (pi/2)[(4sin^2(theta) - 8sin(theta)cos(theta) + 4cos^2(theta))/cos^2(theta)]

B/A = (pi/2)[(2 - 2sin(2theta))/cos^2(theta)]

B/A = (pi/2)(2cos(2theta)/cos^2(theta))

B/A = (pi/4)(sin(2theta)/cos(theta))

Therefore, the ratio of the area of the semicircle to the area of the triangle is (pi/4)*(sin(2theta)/cos(theta)).

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Your statement:

'From A, draw two lines tangent to the semicircle, one to each endpoint of the base of the triangle. Label the points where these lines intersect the base of the triangle B and C. '

Impossible construction!

c------------------------------

As an AI language model, I cannot confirm whether the construction is possible or not without knowing the specific details of the triangle and semicircle in question. However, if the semicircle is completely outside the triangle and does not intersect any of its sides, then it is impossible to draw two lines tangent to the semicircle from the endpoints of the base of the triangle. This is because there would be no point of contact between the semicircle and the lines.

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Only one tangent at any point of arc.

c---------------------------

As an AI language model, I cannot say this statement is true or false without additional context. Could you please provide more details or clarify your intention?

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You will have to learn more math. Farewell.

c---------------------------------------

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