Author |
Topic: Rectangles in triangles (Read 4389 times) |
|
pex
Uberpuzzler
Gender: 
Posts: 880
|
 |
Rectangles in triangles
« on: Feb 12th, 2016, 5:23pm » |
 Quote  Modify
|
What is the largest rectangle that can be inscribed in a given right triangle? I suppose this must be a classic, but I had not seen it before and found the result neat.
|
|
 IP Logged |
|
|
|
markr
Junior Member
Gender:
Posts: 90
|
|
Re: Rectangles in triangles
« Reply #1 on: Feb 12th, 2016, 8:05pm » |
Quote Modify
|
If the lengths of the legs of the triangle are A and B, the dimensions of the rectangle are A/2 x B/2.
|
|
IP Logged |
|
|
|
pex
Uberpuzzler
Gender:
Posts: 880
|
|
Re: Rectangles in triangles
« Reply #2 on: Feb 12th, 2016, 9:02pm » |
Quote Modify
|
on Feb 12th, 2016, 8:05pm, markr wrote:| If the lengths of the legs of the triangle are A and B, the dimensions of the rectangle are A/2 x B/2. |
|
Yes, that is the optimum... for that particular orientation. 
|
|
IP Logged |
|
|
|
pex
Uberpuzzler
Gender:
Posts: 880
|
Well, I found it interesting.
There are only two sensible orientations to consider. One is where two sides of the rectangle are on the legs of the triangle, leading to the solution that markr gave.
The other is where one side of the rectangle is on the hypotenuse of the triangle. In this case, we find an optimal rectangle that is not congruent to the other solution, but does have the same area, covering exactly half of the triangle. Sketch in the attachment.
|
|
IP Logged |
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Rectangles in triangles
« Reply #4 on: Feb 22nd, 2016, 3:57am » |
Quote Modify
|
nice.
|
|
IP Logged |
|
|
|
Hippo
Uberpuzzler
Gender:
Posts: 919
|
|
Re: Rectangles in triangles
« Reply #5 on: Mar 22nd, 2016, 3:50am » |
Quote Modify
|
BTW: If you generalise the problem to parallelograms instead of rectangles, the area ratio would be the same, and affinity simplifies resoning so you can solve it just for equilateral triangle.
From the pex's picture all solutions contain the central black rectangle and one of it's sides is the parallelogram edge. Opposite edge could be moved anyways on the original triangle side.
|
| « Last Edit: Mar 22nd, 2016, 4:17am by Hippo » |
IP Logged |
|
|
|
anglia
Junior Member
Gender: 
Posts: 60
|
|
Re: Rectangles in triangles
« Reply #6 on: Apr 29th, 2016, 11:05pm » |
Quote Modify
|
nice guys...
|
|
IP Logged |
ebooks
|
|
|
rloginunix
Uberpuzzler
Posts: 1029
|
A calculus-free deduction or why half.
Using the right triangle from the drawing by pex, let Y be an arbitrary point on the short and X be the corresponding rectangular point on the long side of the given triangle such that the point of intersection of perpendiculars to the sides of the triangle through Y and X, P, is located on the hypotenuse.
Rotate Y counterclockwise 90 degrees about C until it lands on the extension of CB as Y'. Consider an inversion with respect to a circle in a real plane - the points Y' and X are the inverses of each other with respect to a circle of inversion centered at C with negative(!) power since the center of inversion is located between Y' and X.
The workhorse formula of inversion with respect to a circle is:
y*x = r2
where y and x are the linear distances of Y' and X from C, in our case - the lengths of the sides of the rectangle sought-after, and r is the radius of inversion constructed in ICE4 in the tutorial on this forum modulo negative power of inversion.
Key idea: the variance of two magnitudes is replaced with a variance of a single magnitude only - the length of the radius of inversion which reaches its maximum when the free-floating vertex of the square constructed on r lands on the diagonal of the square whose square area is exactly equal to the square area of the given rectangle, read - right triangle, whose construction is covered in Euclid's Elements B2P14, "to construct a square equal to a given rectilinear figure".
That was my line of thinking, packaging which into a solution we reverse the steps:
1) Given an arbitrary right triangle (rectangle), construct the corresponding (Parent) square of equivalent square area, B2P14, and consider its half, an isosceles right (Parent) triangle DCE
2) Using a geometric transformation of homothety inscribe a square into the Parent (isosceles right) triangle to locate D' on CA
3) Using B6P12 ("to find a fourth proportional to three given straight lines") locate Y on CA since CD, CA and CD' are known. In this case, however, we are lucky since:
CD/CD' = 2
4) Once the location of Y is known, perform a transformation reverse to that of B2P14 - given a square and a magnitude of one side of a rectangle, construct the remaining vertex X on CB:
drawings
(conversely, we may locate X on CB first, by constructing a straight line through D' parallel to DB, and then locating Y on CA by reversing the B2P14 construction)
Comments:
C1) Attached is a GeoGebra design file that I created on or about the date of this posting with GeoGebra 4.2 on a 64-bit OpenSuse 13.2. While on UNIX/Linux the file can be opened as is, if it makes you feel more comfortable you can remove the (artificial) .pdf extension.
Once opened, left-click on Y to move it (orientation and proportions of the given right triangle (C, B) are adjustable).
All the software is "as is", no warranties of any kind.
C2) We observe that the rectangle sought-after and the square with the same square area are the images of each other under the B2P14-style transformation.
C3) Once the problem is solved for one orientation, the second solution follows automatically since the square area of any triangle can be represented as a sum of square areas of two adjacent right triangles each of which we know how to fill with a largest rectangle.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print
rect_in_tri.pdf