computer vision – Detect pentagon in image

Given a UK Energy Performance Certificate (EPC) image, detect the pentagons containing integers between 1 and 100.

(these are isosceles right pentagons)

     img1 = Import("https://i.stack.imgur.com/Jcpnr.png")

enter image description here

enter image description here

Background:

I hope to use a neural net to classify the pentagons in order to parse the EPC certificates. I have tried regular OCR, but this has turned out to be unreliable.

These can be found in the bestiary folder in this github repo.

They can be summed in mathematica via the following code:

path="https://raw.githubusercontent.com/ccosnett/epc_parser/master/assets/bestiary/";
imgs = path <> # & /@ {
                        "102489965-2-ei-rating.png"
                      , "102492044-1-epc.jpg"
                      , "102519884-1-epc-graphs.jpg"
                      , "102729176-2-eir.jpeg"
                      , "102984668-1-epc-rating-graph.png"
                      , "103003823-1-epc.gif"
                      , "103187228-4-epc-rating-graph.png"
                      , "103309151-1-ee-rating.png"
                      , "105825026-1-epc-1.png"
                      , "105825038-1-epc-1.jpeg"
                      , "105829613-1-epc-1.png"
                      , "105835280-1-epc-1.jpeg"
                      , "108300254-1-epc-rating-graph.png"
                      , "108386993-1-environment-impact.jpg"
                      , "108386993-2-energy-efficiency.jpg"
                      , "test-1-12.png"
                      , "test-81-29.png"
                      , "test-88-1.png"};
images = Import /@ imgs

here is what they look like:

enter image description here

I learned from the minimal documentation of the new ImageBoundingBoxes that I can search for Entity‘s in images

The only one I could get to work was Entity("Concept", "StopSign::wp7bc") but this only worked in one of my EPC images 🙁 But this is nevertheless an interesting route to explore.

ImageBoundingBoxes(im1, Entity("Concept", "StopSign::wp7bc"))

enter image description here

Lamina also looked promising, but is unfortunately not a searchable Entity at this time:

enter image description here

FindImageText seems to be too unreliable. And has to be used in conjunction with OCR and some logic to check if it has picked up the numbers on the pentagons:

hh(i_) := HighlightImage(i, FindImageText(#, "Word", AcceptanceThreshold -> 0) &);
hh /@ images

enter image description here

I would really appreciate some ideas 🙂

mg.metric geometry – Collinearity in tangential pentagon

I am looking for a proof of the following claim:

Given tangential pentagon. Touching point of the incircle and the side of the pentagon,the vertex opposite to that side and the intersection point of diagonals drawn from endpoints of that same side are collinear.

enter image description here

The GeoGebra applet that demonstrates this claim can be found here.

mg.metric geometry – Collinearity of three significant points of bicentric pentagon

Can you provide a proof for the following claim?

Claim. Given bicentric pentagon. Consider the triangle whose sides are two diagonals drawn from the same vertex and side of pentagon opposite from that vertex.Then, center of excircle of this triangle which touches side of pentagon, the vertex of pentagon opposite to that side and the incenter of the pentagon formed by diagonals are collinear.

enter image description here

GeoGebra applet that demonstrates this claim can be found here.

geometry – Maximizing the area of a cyclic pentagon made up of triangle centres.

Triangle $ABC$ has $angle BAC=60^circ$, $angle CBA le 90^circ$, $BC=1$, and $AC ge AB$. Let $H$, $I$, and $O$ be the orthocenter, incenter, and circumcenter of $triangle ABC$, respectively. Assume that the area of the pentagon $BCOIH$ is the maximum possible. What is $angle CBA$?

This is problem 25 from the 2011 AMC12A. My friend sent me the following solution:

Angle formulas give that $BCOIH$ is cyclic. Further $I$ and $H$ are both on the arc to the “left” of $O$, with $H$ further “left” than $I$. To maximize the area, we want to distribute $I$ and $H$ evenly over arc $BO$. This gives that $angle{IBC}=angle{IBO}+angle{OBC}=10+30=40$. Multiplying by two gives that the answer is $80$.

I agree with this solution up to the point where they claim they want to distribute evenly abut BO. Why not distribute evenly over the arc BC, doesn’t this make the area even greater? If anyone could explain why we would only distribute on arc BO that would be great.

Inscribed circle in a pentagon

In the figure there is a regular pentagon with a side length of 10 cm. What is the area of the circle? Ignore the fraction and submit the integer value only (if the area is 49.981, submit 49). I know some basic knowledge of trigonometry and geometry but I can’t figure this out.

enter image description here

geometry – Attempt to set color of 2D pentagon to Yellow in Graphics3d fails

When I was working in Graphics, I was just doing the graphic primitive with a color and that was all I needed but when I do it in Graphics3D he fails. I couldn't get the correct syntax for GraphicsComplex. Some suggestions would be appreciated.

.
enter description of image here

pentagram = BoundaryDiscretizeRegion@Polygon@
  CirclePoints[{1, 3 Pi/10}, 5][[{1, 3, 5, 2, 4}]];
ri = RegionIntersection[RegularPolygon[5], pentagram];
Graphics[MeshPrimitives[ri, 1], ImageSize -> 50]
data = Append[#, 0] & /@ 
  Join[PolygonCoordinates[ri], N@CirclePoints[5]]
Graphics3D[{
    FaceForm[], EdgeForm[{Thin, Black}], Yellow,
    Polygon[data, {5, 6, 10, 11, 8}],
    Polygon[data, {10, 11, 15, 14}],
    Polygon[data, {10, 9, 7, 6}],
    Polygon[data, {4, 8, 12, 18}],
    Polygon[data, {3, 5, 1, 19}],
    Black, Text[#, data[[#]] + {-.05, 0, 0}], Point[data[[#]]]
    } & /@ Range[20], Axes -> True]

graphics – Need help creating a pentagon network

I am trying to place 5 pentagons around a pentagon but it is not good. Please help.

enter description of image here

p = Table[{Cos[θ], Sin[θ]} // N, {θ, π/2 + 2 π/5, -3 π/2 + 2 π/5, -2 π/5}]
Graphics[{Line[p],
  Table[Rotate[
    Translate[Line[p],
     1.5 {Cos[t - 18], Sin[t - 18]}],
    t/2],
   {t, 72 Degree Range[5]}]}]

manifolds – Is a pentagon a surface?

My question arises because a pentagon is homeomorphic to a closed disc. The latter is a surface with limits.

However, a pentagon has vertices, so it does not appear to be a double.

If you consider a pentagon without edges, it is a collector at 2, but is not compact, so it is not a surface.

So … any polygon is a surface?

cordially

An equilateral triangle DEF is inside a regular pentagon ABCDE, sharing the vertices D and E. What is the AFC angle?

Confused. Can someone help? I am a math student and I came across this question, but using algebra, I found myself stuck. Did I do something wrong, because my peers could not get it either.

equation triangle relationship of a pentagon

ABCDE is a pentagon of irregular shape. No angle is greater than 180.
prove it,
region of (ABD)(AEC) – domain of (ABC)(ADE) = area of ​​(ABE) * (ACD)