Difference between revisions of "2021 WSMO Team Round/Problem 4"
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Thus, since all the resulting triangles are similar to the one before it, the common ratio of the areas is <math>\frac{3}{4}</math>, and with a infinite series, we get that the summation of the total area is: | Thus, since all the resulting triangles are similar to the one before it, the common ratio of the areas is <math>\frac{3}{4}</math>, and with a infinite series, we get that the summation of the total area is: | ||
| − | <math>\frac{a}{1-q} = \frac{\frac{ | + | <math>\frac{a}{1-q} = \frac{\frac{9\sqrt{3}}{2}}{1-\frac{3}{4}} = \frac{36\sqrt{3}}{2} = 18\sqrt{3}</math> |
The final step is to square this, and we get our desired result, which is <math>(18\sqrt{3})^2 = 972</math> | The final step is to square this, and we get our desired result, which is <math>(18\sqrt{3})^2 = 972</math> | ||
| + | |||
| + | ~akliu | ||
Latest revision as of 10:08, 24 June 2022
Creating a triangle with vertices 
and 
, we immediately notice that triangle 
is a 30-60-90 triangle, with hypotenuse 
, and a longest leg 
.
[Add Image here]
Next, we read the next part of the problem. We create another right triangle similar to the original, but with hypotenuse 
, and legs 
and 
.
[Insert Image here]
Since all the resulting triangles are similar, and we need to find the summation of the area of all of the resulting triangles (to infinity), we must use an infinite geometric series to determine the total area.
The area of the first triangle is 
, and the second triangle (that is also similar to the first triangle), 
.
Thus, since all the resulting triangles are similar to the one before it, the common ratio of the areas is 
, and with a infinite series, we get that the summation of the total area is:
The final step is to square this, and we get our desired result, which is 
~akliu
