2
$\begingroup$

enter image description here

With the details, what could be the value of X?

$\endgroup$

closed as off-topic by Mark Fantini, projectilemotion, GNUSupporter 8964民主女神 地下教會, zoli, Yiyuan Lee Mar 13 '17 at 2:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – projectilemotion, GNUSupporter 8964民主女神 地下教會, zoli, Yiyuan Lee
If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

Let $d(x, y)$ be the distance of two vectors $x, y \in \mathbb{R}^2$. Let $p$ be the point where the two triangles touch. We aim to find $||p||$. $d((15/2, 0), p) = 15/2$ and $d((0, 10), p) = 5$. Let $p = (p_1, p_2)$. Then \begin{align*} \sqrt{(15/2 - p_1)^2 + p_2^2} = 15/2, \ \sqrt{p_1^2 + (10-p_2)^2} = 5 \end{align*} The first gives $p_2 = \sqrt{15 p_1 - p_1^2}$, assuming it is positive. Inserting this into the second we get \begin{align*} \sqrt{p_1^2 + (10-\sqrt{15 p_1 - p_1^2})^2} = 5 \end{align*} And solving for $p_1$ we get $p_1 = 3$, so that $p_2$ must be $6$, using the other equations.

$||p||$ must therefore be $\sqrt{ 3^2 + 6^2} = \sqrt{45} = 3 \sqrt{5}$

More detail upon request.


Edit: it has been pointed out by Andrei in the comments below that the equations that solve for the coordinates of $p$ can be geometrically interpreted as finding the point at the intersection of two circles.

$\endgroup$
  • 3
    $\begingroup$ Geometrical interpretation: the locus of points where a segment is seen at $90^\circ$ is a circle centered on the center of the segment, with the radius half of the segment length. $\endgroup$ – Andrei Mar 12 '17 at 19:13
  • $\begingroup$ $p_2=6$, not $5$. $\endgroup$ – Aretino Mar 12 '17 at 19:29
  • $\begingroup$ My bad. Edits made. $\endgroup$ – Dean Young Mar 12 '17 at 19:36
  • $\begingroup$ @Andrei So you're looking for the intersection of two circles. There might not be any, or there might be two. In this case it seems as if there's just one, so the circles are tangent. I think you should write this up as another answer. $\endgroup$ – Ethan Bolker Mar 12 '17 at 19:44
  • $\begingroup$ @EthanBolker It is the same answer as above, I just wanted to clarify where did the equations Dean Young wrote at the beginning came from $\endgroup$ – Andrei Mar 12 '17 at 20:16
2
$\begingroup$

Let the hypotenuse of both triangles be the diameter of two circles: enter image description here

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.