The Ice Ocean Cathedral
If the front of the cathedral has the shape of an isosceles triangle and the base is 5/6 of the height and the perimeter is 55m and the temperature is freezing how tall is the church?
Problem source: Mattenøtt, NRK.
If the front of the cathedral has the shape of an isosceles triangle and the base is 5/6 of the height and the perimeter is 55m and the temperature is freezing how tall is the church?
Problem source: Mattenøtt, NRK.
March 12th, 2010 at 3:38 am
Let h = height of the church.
The base, therefore = 5h/6, as does each side.
3•5h/6 = the perimeter = 55 meters.
5h/2 = 55
5h = 110
h = 22 meters
March 12th, 2010 at 4:10 am
Then it wouldn’t be equilateral. Let’s suppose that, like the buildings in the pictures, it’s isosceles, and the sloping sides are equal. Let the base be 10x. Then the height is 12x. By Pythagoras’s Theorem, each sloping side is the hypotenuse of a right-angled triangle whose legs are 5x and 12x, so it is 13x. Then the perimeter is 2*13x+10x=36x, which is 3 times the height. This perimeter is 55m, so the height is 55/3 m = 18 1/3 m.
March 12th, 2010 at 4:12 am
Sorry: I meant that comment “Then it wouldn’t be equilateral.” to be a response to the phrase “the base is 5/6 of the height” in the statement of the problem.
March 12th, 2010 at 6:02 am
Sorry, I got the terms mixed up. I wrote equilateral, but meant isosceles. It has now been corrected in the statement of the problem.
March 12th, 2010 at 9:04 am
Ahh well thank you. That certainly changes the approach. I hadn’t even considered checking to ensure that b = 5h/6 was correct for an equilateral triangle.
x = length of a side. h = height. 5h/6 = base.
sides in terms of h: ?[(5h/12)² + h²] = ?[25h²/144 + h²] = ?[169h²/144] = 13h/12
55 = 26h/12 + 5h/6 = 36h/12 = 3h
55/3 = h
Thus I concur with Richard. =)