0
$\begingroup$

If you have a square inscribed inside of a right angled triangle. Call the sides of the square $k$. The hypotenuse is $z$. How would you express either of the two other sides of the triangle in terms of $z$ and $k$? The corner of the square touches the hypotenuse-there isn't a side of the square against the hypotenuse.

$\endgroup$
1
  • $\begingroup$ Does another corner of the square coincide with the right angle of the triangle? $\endgroup$
    – Henry
    May 30, 2020 at 20:15

2 Answers 2

0
$\begingroup$

Hint:

Call one angle of the right triangle $\theta$. The square produces two more smaller right triangles whose hypotenuses add up to $z$, i.e. $$k\sec\theta+k\csc \theta =z \\ \frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=\frac zk \\ \frac{1+2t}{t^2}=\frac{z^2}{k^2}$$ where $t=\sin\theta \cos\theta$. You can solve for $t$ and consequently $\theta$ using this equation. Once you have $\theta$, the required sides will be $k+k\tan\theta$ and $k+k\cot\theta$.

$\endgroup$
9
  • $\begingroup$ I'm not sure how to solve for t and theta. Could you provide this solution? $\endgroup$
    – tomm0334
    May 30, 2020 at 20:37
  • $\begingroup$ @tomm0334 Can you solve the quadratic equation for $t$? You have to remember to take the positive root. After you have obtained $t$, $$\sin\theta \cos\theta =t \implies \frac 12 \cdot 2\sin\theta\cos\theta=t \implies \sin 2\theta =2t \implies \theta =\frac 12 \sin^{-1}(2t)$$ $\endgroup$
    – Vishu
    May 30, 2020 at 20:45
  • $\begingroup$ I get a math error. I got t=4. When I put it in, I get math error on the calculator. $\endgroup$
    – tomm0334
    May 30, 2020 at 21:06
  • $\begingroup$ @tomm0334 $t=4$ is impossible. Why don’t you try calculating it by hand? $\endgroup$
    – Vishu
    May 31, 2020 at 9:47
  • $\begingroup$ I don't understand how to solve by hand? $\endgroup$
    – tomm0334
    May 31, 2020 at 10:20
0
$\begingroup$

Hint:

If the other sides are $x$ and $y$ then you have $$\sqrt{x^2+y^2}=z$$ $$\sqrt{(x-k)^2+k^2} +\sqrt{(y-k)^2+k^2} =z$$

which are two equations in the two unknowns. You should expect multiple solutions and so should check for spurious solutions introduced by squaring

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .