Create a triangular image with height of 200 pixels and the base of 200 pixels and color the image in any two colors.

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Cesar 2013-2-13

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Image Analyst 2013-2-13

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Hint: use poly2mask() and logical indexing.

yourImage = zeros(rows, columns, 'uint8'); % Initialize.
% Assign values to triangle defined by "mask" image.
yourImage(mask) = someValue;

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Image Analyst 2013-2-15

编辑：Image Analyst 2013-2-15

在 MATLAB Online 中打开

You got your 2 color triangle like you wanted. The left triangle is just the mask I made with poly2mask(). Here, I've put in some title's to explain that:

color1 = uint8([100 150 200]);
color2 = uint8([180 50 250]); 
rows = 480;
columns = 640;
redChannel = color1(1) * ones(rows, columns, 'uint8');
greenChannel = color1(2) * ones(rows, columns, 'uint8');
blueChannel = color1(3) * ones(rows, columns, 'uint8');
% Get/assign triangle vertices, triangleX, triangleY.
% I'm sure you can figure out what goes here!
triangleX = [0 100 200]; 
triangleY = [0 200 0]; 
% Create mask.
mask = poly2mask(triangleX, triangleY, rows, columns);
% Display mask
subplot(1,2,1);
imshow(mask);
title('This is the mask poly2mask made', 'FontSize', 25);
% Assign values to triangle defined by "mask" image.
redChannel(mask) = color2(1);
greenChannel(mask) = color2(2);
blueChannel(mask) = color2(3);
rgbImage = cat(3, redChannel, greenChannel, blueChannel);
subplot(1,2,2);
imshow(rgbImage);
title('This is your triangle', 'FontSize', 25);
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
% Give a name to the title bar.
set(gcf,'name','Demo by ImageAnalyst','numbertitle','off')

If you want, you can just get rid of the first subplot() and imshow(). This solves your problem, so go ahead and mark it "Accepted."

Walter Roberson 2013-2-15

Cesar, you need to look more carefully at the image you supplied. Zoom in on the area near the tip. You will notice a few things:

There appear to be three tips. The left tip is white, the center tip is white, the right tip is dark.
The dark tip is part of a dark line that is most easily explained as if the dark line is the shadow of the center tip with respect to a light source that is to the left of center of the image.
If the left (white) tip and center (white) tip were at equal heights, then both would have shadows, but there is only one shadow
The details of the edge crossings of the left and center tips strongly suggest the interpretation that the left triangle is on top of the right triangle, with the left triangle ending a small bit to the right of the right triangle.
the bottom line of the triangles show that the triangles continue to be overlapped all the way down