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# Triangle

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An isosceles triangle has two sides of length $7$ and an area of $14.$ What is the product of all possible values of its perimeter?

Feb 9, 2024

#1
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We see trigonmetry might be a possible solution to the problem, but here is a non-trigonmetry solution.

Instead, we focus on the fact that this triangle is an isosceles triangle, so we can draw a perpendicular to solve this problem.

Consider the diagram below.

We can now use a system of equations to solve this problem. $$\sqrt{{x}^{2}+{y}^{2}}=7, \frac{2xy}{2}=14$$

Simplifying, we get $${x}^{2}+{y}^{2}=49, xy=14$$. This is not an easy system to solve, but there is a small trick I often use for a problem like this. We recognize the structure of $$x+y=\sqrt{{x}^{2}+2xy+{y}^{2}}=\sqrt{49+14*2}=\sqrt{77}$$. Now we can use a little trick with the Vieta's formulas. We recognize that x and y are the two roots of the equation $${a}^{2}-\sqrt{77}a+14$$. (because the sum of the roots is sqrt(77) and the product of the roots is 14). We recoginze here, that x and y are interchangable, because we can assign any of the roots to x or y.

But we need to read the problem. It is not asking the roots, but the product of the perimeters. Because we know x and y are the two roots, or the two solutions to the base, we calculate $$(14+2x)(14+2y)=196+28(x+y)+4xy=196+28\sqrt{77}+56=252+28\sqrt{77}$$. So our answer is 252+28sqrt(77).

Feb 10, 2024
edited by hairyberry  Feb 10, 2024
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Clever solution hairy! If you don't know trig, this is the way to go-

proyaop  Feb 10, 2024
#2
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The area of a triangle is $$ab\sin{C}\over{2}$$, along with b*h*1/2, but I prefer the first one for this problem. (You can look the proof up, it is pretty simple)

We know that a = b = 7, so the area of the triangle = 49sinC/2 = 14. Then solving for sinC = 4/7.

Because we want to find the third length, it is best to use the Law of Cosines for C, but how?

Using the identity $$\sin^2\theta+\cos^2\theta=1$$, where $$\theta=C$$, we know that $$\cos^2C=1-\sin^2C=1-({4\over7})^2=1-{16\over{49}}={33\over{49}}$$. Square rooting both sides, you get $$\cos{C}=\pm{\sqrt{33}\over{7}}$$. Now, we can effectively do Law of Cosines: c^2 = a^2 + b^2 - 2ab*cosC.

We will do 2 seperate calculations of c from law of Cosines based on the two possible values of cosC.

Case 1: $$\cos{C} = +{\sqrt{33}\over{7}}$$, keep in mind that c is the third side length we don't know, and a = b = 7.

$${c_1}^2=7^2+7^2-2*7*7*{\sqrt{33}\over{7}}=98-14\sqrt{33}$$, and since c has to be positive to be a real triangle:

$$c_1 = \sqrt{98-14\sqrt{33}}$$

Case 2: $$\cos{C}=-{\sqrt{33}\over{7}}$$; a = b = 7 still.

$${c_2}^2=7^2+7^2-2*7*7*-{\sqrt{33}\over{7}} = 98 + 14\sqrt{33}$$, which yields

$$c_2=\sqrt{98+14\sqrt{33}}$$, (again, there is no plus/minus because c has to be positive to be a real triangle.)

Setting c1 = a + b*sqrt(c) is a great strategy for radical simplification, and then you can square both sides and solve system of equations (with substitution, and yes, I did it)! Oh and by the way, don't expect a good answer for this problem...

After simplification, you should reach $$c_1=\sqrt{77}-\sqrt{21}; c_2=\sqrt{77}+\sqrt{21}$$

Product of 2 possible perimeters = (7 + 7 + c1)(7 + 7 + c2) = [14 + sqrt(77) - sqrt(21)][14 + sqrt(77) + sqrt(21)], and you can do difference of squares: a^2 - b^2 = (a + b)(a - b), where a = 14 + sqrt(77) here, and b = sqrt(21):

= 252 + 28sqrt(77)

Hence, our final answer is $$252 + 28\sqrt{77}$$

Feb 10, 2024