B Couple geometry/trigonometry questions

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The discussion revolves around a question from Gelfand's Trigonometry regarding the circumradius of a triangle with sides 6, 8, and 10. It confirms that the hypotenuse of a right triangle inscribed in a circle must indeed be the diameter. Participants clarify the original question, noting it refers to a circle drawn through the triangle's vertices, not its sides. The solution to the circumradius question is determined to be 5. The thread also touches on the difficulty of finding solutions to Gelfand's book, but no specific resources are provided.
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I am reading Gelfand's Trigonometry. In one of the questions he asks: "We know from geometry that a circle may be drawn through the three vertices of any triangle. Find the radius of such a circle if the sides of the triangle are 6,8, and 10."

My first question is, I know that if the diameter of a circle is the hypotenuse of a triangle then that triangle is a right triangle. Does this imply that the hypotenuse of any right triangle inscribed within a circle must be the diameter?

If this is not the case then I'm at a loss on how to solve this problem.

Second question, I have searched around but cannot find solutions to this book is there a place to find the solutions.
 
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Velcroe said:
I know that if the diameter of a circle is the hypotenuse of a triangle then that triangle is a right triangle. Does this imply that the hypotenuse of any right triangle inscribed within a circle must be the diameter?
Yes, it must be the diameter.
 
Velcroe said:
I am reading Gelfand's Trigonometry. In one of the questions he asks: "We know from geometry that a circle may be drawn through the three sides of any triangle. Find the radius of such a circle if the sides of the triangle are 6,8, and 10."

My first question is, I know that if the diameter of a circle is the hypotenuse of a triangle then that triangle is a right triangle. Does this imply that the hypotenuse of any right triangle inscribed within a circle must be the diameter?

If this is not the case then I'm at a loss on how to solve this problem.

Second question, I have searched around but cannot find solutions to this book is there a place to find the solutions.
Unfortunately I cannot answer your last question. The answer to the first, however, is yes. Imagine you have the hypotenuse of a right triangle in a circle and it is not the diameter. Then for the third point to be on the circle you get either a longer side which cannot be true or an angle which cannot be right which cannot be true either.
 
fresh_42 said:
Unfortunately I cannot answer your last question. The answer to the first, however, is yes. Imagine you have the hypotenuse of a right triangle in a circle and it is not the diameter. Then for the third point to be on the circle you get either a longer side which cannot be true or an angle which cannot be right which cannot be true either.

That makes sense, so the answer to Gelfand question quoted above would just be 5. Seems like the question is too easy which is why I asked my question in the first place. Well thank you for your response.
 
Velcroe said:
"We know from geometry that a circle may be drawn through the three sides of any triangle.
I think the author means a circle inscribed in a triangle, instead of the opposite. He said "sides", not "corners".
Velcroe said:
I have searched around but cannot find solutions to this book is there a place to find the solutions
If assuming the circle inscribed in a triangle is correct, then will this animation be helpful?
 
Actually my mistake in quoting the question. It actually states"... may be drawn the the three vertices of any triangle". Sorry about that don't know how I mistyped that. Fixed my original question.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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