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- Jul 17th 2018, 12:43 PM #1

- Jul 17th 2018, 12:50 PM #2

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- Jul 17th 2018, 02:17 PM #3
## Re: problem

- Jul 17th 2018, 05:02 PM #4

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- Jul 17th 2018, 06:11 PM #5
## Re: problem

@Denis, two angles cannot be equal. If $\angle ACB=\angle AMC$ then that is only one angle.

But $\angle ACB=\overrightarrow {CA} \cup \overrightarrow {CB}$**AND**$\angle AMC=\overrightarrow {MA} \cup \overrightarrow {MC}$ which shows that those angles cannot be the same angle.

Now it is true that $m(\angle ACB)=\dfrac{\pi}{2}=m(\angle AMC)$

- Jul 18th 2018, 05:35 AM #6

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- Jul 18th 2018, 08:46 AM #7

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## Re: problem

I'll refer to Plato's diagram. Because Triangle ACB is a right triangle, and Line segment CM is perpendicular to AB, and meets Triangle ACB at vertex C

(which is across from the hypotenuse), then this forms three right triangles, all of which are similar to each other:

Triangle AMC, Triangle MCB, Triangle ACB

Let a = measure of Line segment AC

Let b = measure of Line segment CM

Let c = measure of Line segment CB

I will look at the respective ratios of the shorter legs to the hypotenuses, in order from smaller to larger areas of triangles:

$\displaystyle \dfrac{9}{a} \ = \ \dfrac{b}{c} \ = \ \dfrac{a}{25}$

$\displaystyle \dfrac{9}{a} \ = \ \dfrac{a}{25} \ \implies \ a \ = \ 15$

You might see the value of b right away as being a multiple from the common 3, 4, 5 Pythagorean Triple.

Or, use a version of the Pythagorean Theorem to find it. (b = 12)

Then you could use this to solve for c:

$\displaystyle \dfrac{b}{c} \ = \ \dfrac{a}{25} $

$\displaystyle \dfrac{12}{c} \ = \ \dfrac{15}{25} $

$\displaystyle \dfrac{12}{c} \ = \ \dfrac{3}{5} $

$\displaystyle \cdots$

- Jul 18th 2018, 09:00 AM #8

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## Re: problem

I used the Pythagorean Theorem three times to get:

$$\begin{align*}b^2+9^2 & = a^2 \\ b^2+16^2 & =c^2 \\ a^2+c^2 & = (9+16)^2\end{align*}$$

Let $x=a^2, y=b^2, z=c^2$. This gives:

$$\begin{align*}y+81 & = x \\ y+256 & = z \\ x+z & = 625\end{align*}$$

And from here, we can easily solve the system of three equations:

Plugging in for $x$ from the first equation into the last equation gives:

$$y+z=544$$

Added to the middle equation $z-y=256$ gives:

$$2z=800 \Longrightarrow z=400$$

$$400+y=544 \Longrightarrow y=144$$

$$x=144+81 \Longrightarrow x = 225$$

Taking square roots gives the lengths of the sides, as well.

- Jul 18th 2018, 10:24 AM #9

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- Jul 18th 2018, 10:49 AM #10
## Re: problem

@All, note that is question was on an 1869 admission test for MIT.

I think that the question was designed to test the use of the.**mean proportional theorem**

Recall that in 1869 these answers would require the showing of all work.

The grader would have been look for certain methods.

- Jul 18th 2018, 12:29 PM #11

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