0
$\begingroup$

Note: I am paraphrasing this problem

Consider a quadrilateral with 3 sides of equal length, and one longer side. This quadrilateral also has equal diagonals, both of which are equal in length to the longest side of this quadrilateral. The longer and shorter sides add up to 180 in length. Find the longest side of the quadrilateral.

What I have so far:

  1. The quadrilateral is an isosceles trapezoid

I've been working on this problem a couple of days. I've tried extending the trapezoid into a triangle. I've also tried to make a system of equation with the lengths of the long and short sides of the trapezoid. Neither of these worked. I figure this could somehow be solved with trigonometry, but I don't know exactly how. Does anyone have any ideas?

$\endgroup$
  • $\begingroup$ Generally not a good idea to tell us in the title that it is a "tough" problem. The title is used by many people to figure out if they want to read the question. Give us an idea what the question is about. $\endgroup$ – Thomas Andrews Jan 5 '14 at 18:38
  • $\begingroup$ Do you have any ideas for a better title? I just took the word tough out of it, but I can't think of a better title than that $\endgroup$ – recursive recursion Jan 5 '14 at 18:41
  • $\begingroup$ Yes, here you have to stroke the ego of people that have huge egos. You can get your question closed by people thinking: "How dare this guy call his problem tough when I have solved some many problems tougher than that." The truth is, you are not really the one in fault. You are honestly saying you find it tough. But people with faulty characters, and egos team up. $\endgroup$ – user119256 Jan 5 '14 at 18:43
  • $\begingroup$ I kinda figured it would make them try harder to solve the problem, but if what you're saying is true, I guess it does the opposite. $\endgroup$ – recursive recursion Jan 5 '14 at 18:46
7
$\begingroup$

Hint 1. Add one more point to get whole thing being a regular pentagon.

Hint 2. Remember golden ratio?

$\endgroup$
  • $\begingroup$ Good eye! I'm kicking myself for missing it at first. $\endgroup$ – David H Jan 5 '14 at 18:55
  • 1
    $\begingroup$ Oh believe me, I'm kicking myself harder. $\endgroup$ – recursive recursion Jan 5 '14 at 19:00
  • $\begingroup$ I like your answer better than mine. Great eye $\endgroup$ – user44197 Jan 5 '14 at 21:00
2
$\begingroup$

One of the angle at the base has $$ \cos(\theta)={{\sqrt{5}-1}\over{4}}$$

Recognize the angle?

I found it by 1) Normalizing the three sides to length 1 2) Set one of the vertices at (0,0) and the other at (1,0) 3) If the base angles are $\theta$ and $\phi$ set up the equation in terms of $\theta$ and $\phi$. 4) Show that $\theta=\phi$ and solve.

I am working on a figure...will paste here soon.

Here is the figure enter image description here

$\endgroup$
  • $\begingroup$ How did you find the angles of the trapezoid? $\endgroup$ – recursive recursion Jan 5 '14 at 19:32
  • $\begingroup$ See Gina's answer. I like his/hers better! $\endgroup$ – user44197 Jan 5 '14 at 21:01
  • $\begingroup$ Wow, I officially declare her a genius $\endgroup$ – recursive recursion Jan 5 '14 at 23:22
  • $\begingroup$ I agree. she is! $\endgroup$ – user44197 Jan 6 '14 at 0:04
1
$\begingroup$

Also hints (the other answers are better, this one might just expand your toolbox - useful for this type of questions).

Note that your quadrilateral (isosceles trapezoid, as you said) is also inscribable.

  1. Law of cosines (for any triangle) http://en.wikipedia.org/wiki/Law_of_cosine
  2. Ptolemy's theorem (for inscribable quadrilaterals) http://en.wikipedia.org/wiki/Ptolemy%27s_theorem
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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