The third problem of the week is now up at the big website. I've also posted the official solution to Problem Two. So go have a look and let me know what you think. Feel free to present solutions in the comments.
Unless, of course, you just want to pick micronits with the problem statement. In that case you don't need to let me know what you think.
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480
Using the diagram
B = altitude = 30
Since perimeter = 100, 2a + 2c =100, then a + c = 50
A = 50-C
Using a^2 + b^2 = C^2, and b=30
(50-c)^2 + 900 = c^2
Solving for c, c=34
Solving for a, a = 50 - 34 = 16
Area of triangle = bh/2
Base = 2a = 32
Height = b = 30
Area = 32* 30/2
Area = 480
?
The altitude divides the isosceles triangle into two congruent triangles. This is proven by the AAS theorem - the angles formed by the altitude are right angles, the angles opposite the congruent sides are congruent and the congruent sides are congruent. Therefore, the base of the triangle is bisected by the altitude. Let a be equal to the length of the congruent sides, and let 2b be equal to the base. Then the perimeter is 2a + 2b = 100 or a+b=50. The area is 1/2*2b*30 or 30b. Each of the two congruent triangles is a right triangle with hypotenuse a and legs 30 and b. Therefore, a^2 = b^2 + 900. Substituting 50-b for a gives (50-b)^2 = b^2 + 900. Expanding the left side gives 2500 - 100b +b^2 = B^2 + 900. Combining like terms gives -100b = -1600 or b =16. The area is 30b, so the area must be 480.
jrosenhouse wrote (February 9, 2015):
> The third problem of the week is now up
> An isosceles triangle has a perimeter of 100.
The altitude drawn from the common vertex of the two congruent sides has length 30.
Find the area of the triangle.
The area of the described triangle is
equal to
the area of a rectangle with
one pair of sides having length 30, and
the other pair of sides having length
((100/2)^2 - 30^2) / (2 * 100/2) = 1600 / 100 = 16.
Likewise:
The area of the described triangle is
equal to
the area of 480 squares, all of side lengths 1.
> Unless, of course, you just want to pick micronits with the problem statement. In that case you don’t need to let me know what you think.
Sorry, could not resist.
If this were an SAT or GRE problem, you would note the round numbers and go through the Pythagorean triples you had to memorize to look for a factor of 30. Noting that 30:40:50 doesn't match the perimeter requirements, you would settle quickly on 8:15:17, verify that 8+17=25, then multiply 8•15=2 hours = 120, and then you'd remember that doubling the dimensions quadruples the area. 480 it is.
@4: I did something like that. I made an initial error which caused me to calculate the hypotenuse to be some irrational number (or at least, very long decimal). I immediately rejected that as not a likely solution, then figured out what I had done wrong and found the correct whole numbers. Meta-math - using (likely) info about the problem set-up to help solve the problem. :)
Similar to Sean T's solution, but with more algebra.
The altitude divides the isosceles triangle into two right triangles, with sides a (the known altitude), b (unknown side) and h (unknown hypotenuse).
It is known that the altitude bisects the base of an isosceles triangle. Call the base of the isosceles triangle 2*b, with b being the unknown side of the right triangle as above. Each side of the isosceles triangle is the hypotenuse h of the corresponding right triangle.
As the two right triangles formed by the altitude are congruent the perimeter is 2*h + 2*b, or h+b is the sum of the lengths of the unknown side and the hypotenuse of the right triangles. Call this sum q.
Thus we have two equations with two unknowns (h,b):
[1] h^2 = a^2 + b^2
[2] h + b = q
Solve [2] for h:
h = q - b
Square it:
h^2 = q^2 - 2*b*q + b^2
Substitute into [1]:
q^2 - 2*b*q + b^2 = a^2 + b^2
which simplifies to:
[3] b = (q^2 - a^2)/(2*q).
Since a = 30 and q = b + h = 100/2 = 50
b = (2500 - 900)/100 = 16.
For a right triangle with sides 16 and 30, the area of the right triangle is A = a*b/2 = 30*16/2 = 240.
As the isosceles triangle in the original problem is formed of two triangles of area 240, the total area of the isosceles triangle is 480.
Also, since q = h + b, so h = q - b, or h = 50 - 16 = 34.
Of course, the official solution will be along the lines of what azoomer and Sean T suggest. But this problem, along with several of the others I will be using, came from old high school math competitions I have in my files. It was fairly common solve problems in the manner of what Another Matt writes. That is, we would assume that the answer would work out to something elegant and go from there.
I did it similar to azoomer, but I took as a hint the form for the Pythagorean theorem, and noted that b^2 = (c+a)(c-a), and that 2c+2a = 100, so c+a = 50, yielding 900 = 50(c-a), or c-a = 18.
Given c+a=50 and c-a=18, it's easy to add the two equations to get 2c = 68; c=34 and to subtract the two equations to get 2a=32; a=16.
The area is then A = 2a*b/2 = 16*30 = 480.
Here is a real-world problem I ran into recently that can be easily solved using the two-chord power theorem (or Pythagorean Theorem): You have spheres of diameter d, and some sheets of plywood of uniform thickness t. You want to drill holes in the plywood so that the spheres rest in the holes. When you put a piece of plywood on a table, the holes must not be so large that the spheres touch the table. What is the largest hole you can drill in the plywood?
Hi George. You posted your comment while I was focused on preparing for my trip, so I've only just had a chance to think about it. It's a little hard to describe without drawing a picture, but it seems to me the solution is this: Imagine the sphere resting in the hole we just drilled in the wood. If we take a vertical slice through the sphere, through the prime meridian, the cross-section will be a circle with two chords. One chord will be the diameter of the sphere. The second chord is along the plane of the wood, and represents a diameter of the hole we just drilled. Now, the first chord is divided into pieces of length t and d-t. The second chord is bisected into two segments of length r, where r represents the radius of the hole. What we are looking for, in other words. The two chord theorem now tells us that r^2=t(d-t), from which the answer follows immediately.
How's that?
Jason, that is perfect! I will send you an article I wrote where this formula is applied in the construction of a puzzle.