Difference between revisions of "1992 OIM Problems/Problem 5"

Line 5: Line 5:
  
 
== Solution ==
 
== Solution ==
 +
 +
[[File:1992_OIM_P5a.png]]
 +
 +
First we realize that the trapezoid is isosceles because it is inscribed in a circle.  If we draw a rhombus using the midpoints of the trapezoid, the diagonals of the rhombus will be <math>m/2</math> and <math>h</math>  We can construct such rhombus.  Then, with similar triangles and the midpoints we know that the diagonal of the trapezoid is twice the length of the rhombus.  Now we have all we need to start constructing:
 +
 +
 
* Note.  I actually competed at this event in Venezuela when I was in High School representing Puerto Rico.  I'm proud to say that I got full points on this one and I solved it very quickly.  I had a straight rule and compass kit which I used to solve it as we're supposed to build the trapezoid with it.  Now, 3 decades later, I attempted this and spent a full 3 hours on it and couldn't solve it nor I remember what I did.  I will attempt again some other time.
 
* Note.  I actually competed at this event in Venezuela when I was in High School representing Puerto Rico.  I'm proud to say that I got full points on this one and I solved it very quickly.  I had a straight rule and compass kit which I used to solve it as we're supposed to build the trapezoid with it.  Now, 3 decades later, I attempted this and spent a full 3 hours on it and couldn't solve it nor I remember what I did.  I will attempt again some other time.
{{solution}}
+
 
 +
{{alternate solutions}}
  
 
== See also ==
 
== See also ==
 
https://www.oma.org.ar/enunciados/ibe7.htm
 
https://www.oma.org.ar/enunciados/ibe7.htm

Revision as of 23:42, 19 December 2023

Problem

The circumference $C$ and the positive numbers $h$ and $m$ are given so that there is a trapezoid $ABCD$ inscribed in $C$, of height $h$ and in which the sum of the bases $AB$ and $CD$ is $m$. Build the trapezoid $ABCD$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

1992 OIM P5a.png

First we realize that the trapezoid is isosceles because it is inscribed in a circle. If we draw a rhombus using the midpoints of the trapezoid, the diagonals of the rhombus will be $m/2$ and $h$ We can construct such rhombus. Then, with similar triangles and the midpoints we know that the diagonal of the trapezoid is twice the length of the rhombus. Now we have all we need to start constructing:


  • Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. I'm proud to say that I got full points on this one and I solved it very quickly. I had a straight rule and compass kit which I used to solve it as we're supposed to build the trapezoid with it. Now, 3 decades later, I attempted this and spent a full 3 hours on it and couldn't solve it nor I remember what I did. I will attempt again some other time.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

https://www.oma.org.ar/enunciados/ibe7.htm