<p>Problem from Barron's 2400 (2008 Ed.) "Topics in Geo" (p.299/306)</p>
<p>Too lazy to scan in the problem, but it can be drawn out from the description.</p>
<p>The triangles inside triangle ABC are formed by joining the midpoints of the sides and then repeating the process. If a point is chosen at random inside triangle ABC, what is the probability that the the point lies in the shaded region. (There is a big triangle, a smaller one, a smaller one, and yet a smaller one. The smallest one is shaded.)</p>
<p>The answer is 1/64. I get most of their explanation, but I just don't understand why the second biggest triangle (DEF in the book) is 1/4 the area of triangle ABC.</p>
<p>Thanks! Sorry that I'm way too lazy to scan the diagram!</p>
<p>Take Triangle ABC. If each side of ABC is of length 1, then each side of DEF (formed from the midpoints) is 1/2.</p>
<p>The base of ABC is 1. If ABC is equilateral, then its height is (sqrt3)*(b)/2 which is (sqrt 3)/2. This is because cutting ABC in half vertically makes 30-60-90 triangles.</p>
<p>The base of DEFis 1. If DEFis equilateral, then its height is (sqrt3)<em>(b)/2 which is (sqrt 3)</em> (1/2)/2 which is (sqrt3)/4. This is because cutting ABC in half vertically makes 30-60-90 triangles.</p>
<p>Thus, Area of ABC = (1/2)(1)(sqrt3)/2
Area of DEF = (1/2)(1/2)(sqrt3)/4</p>
<p>The area of DEF is 1/4ths of the area of ABC. This repeats itself. You could also draw out 3 other triangles within ABC that are congruent to DEF and it would fit perfectly if drawn to scale. Basically, it's just an easier thing to memorize for the SAT.</p>
<p>Is it an equilateral triangle? If it is, then connecting the midpoints creates four congruent triangles: You are creating similar triangles with a scale factor of 2 with respect to the large triangle. So the area of each is one-fourth. You can also determine that all the triangles are congruent, which implies that they all have equal area (1/4 of the original).</p>
<p>Midline (midsegment) Theorem: The segment connecting the midpoints of two sides of a triangle is parallel to the third side and half the length of that side.</p>
<p>Thus joining all three midpoints breaks ANY triangle into four congruent triangles.
Let's call the triangle formed by midsegments midtriangle.
If the area of ABC is 1, then the area of its midtriangle is 1/4;
the area the midtriangle's midtriangle is (1/4)/4 = 1/16;
the area of the next midtriangle is (1/16)/4 = 1/64.</p>
<h1>Probability = 1/64.</h1>
<p>Google the similarity ratio; knowing its properties helps in dealing with ratios of areas and perimeters of similar triangles.</p>
<h1>A triangle and its midtriangle are similar with the similarity ratio 1/2; their areas ratio is 1/4.</h1>
<p>Here's a good question:
The midquadrilateral is formed by joining the midpoints of the convex quadrilateral's consecutive sides. What's the ratio of the areas of the big quadrilateral and the midquadrilateral?</p>