A Japanese Temple Geometry Problem from 1800.

here we're going to look at a nice japanese geometry problem so this comes from a sengaku from the ida school in about 1800 and so these were geometry problems that were drawn onto wooden blocks or i should say painted onto wooden blocks and so here's the setup we've got a circle of radius r and inside of that circle we have an equilateral triangle which has been inscribed and then at the midpoint of one of the sides of the equilateral triangle we put another equilateral triangle with two vertices on the circle and one at that midpoint and our goal is to find q which is the length of the side of this smaller equilateral triangle in terms of r which is the radius of the circle so our main strategy here will be to add some line segments into this picture and then use trigonometry so the first thing that i want to do is put a center to this circle so maybe i'll put a center right here and i'll call the center o for like the origin then next i want to draw two radii of this circle so i'm going to draw a radii from here to this vertex of this triangle so of our larger equilateral triangle so we know that has length r and i'm going to draw one more from here down to the vertex of this smaller triangle so again we know that this also has length r good then the next thing that i want to do is notice that we know this angle measure this angle measure is going to be 30 degrees or pi over 6. so let's maybe talk our way through that really quickly so we can also put a line segment here so i will dot it just because it's not as important as these others and we can put a line segment here and then let's look at these three triangles that we've used to split up the equilateral triangle and notice that they are all congruent by the side-side-side theorem they all share one side of the bigger triangle but we know that's an equilateral triangle and then they share these line segments in here which all have length r so in other words this triangle right here this one right here and this one right here are all congruent but that means we've got two angles right here that are equal in measure and add up to 60 degrees which is the measure of the angle of an equilateral triangle so that means this is half 60 degrees or 30 degrees or pi over 6. good now the next thing that i want to do is add a line segment here from the origin down to this point right here which is the midpoint of our larger equilateral triangle and the vertex of this smaller equilateral triangle now i'm going to go ahead and call this length x and we'll calculate x over there in just a second then i'm going to also extend this thing down to the midpoint of the base of the smaller equilateral triangle and i'm going to call this length right here y and also notice i know the angle measure of this angle because this is an equilateral triangle that's going to be pi over 3. good now we can use trigonometry to calculate the length of x and y in terms of r and q given the fact that we've got right triangles here with hypotenuse r in this case hypotenuse q in this case height x in this case and height y in this case so let's go ahead and notice that we have x is equal to r times the sine of pi over six but notice that's equal to r over two sine pi over six is half and then we also see that y is equal to q times the sine of pi over 3 but that's equal to q times the square root of 3 over 2 because sine pi over 3 is root 3 over 2. okay great and now we have one thing left to notice and that is we've got this larger right triangle which i'll maybe shade in red now so we've got this length right here this length right here and then the length right here which was purple and we know the length of all of the sides of that right triangle this one down here is q over two because we have the midpoint right here of this segment which is the base of the equilateral triangle so it's half the length of the whole thing so that tells us we've got a right triangle where the sum of these is the length of one of the sides notice that's x plus y and so we have r over two plus q times root three over two so this is the length of one side and then the length of another side is q over two so let's maybe put here these are side lengths and then we know the hypotenuse has length r which is the radius of the circle so maybe we'll put that as hypotenuse length so the next thing that we want to do is apply the pythagorean theorem to the triangle which i drew over there in red that has those two side lengths and that hypotenuse length so let's see what that gives us so we have r over 2 plus q times root 3 over 2 quantity squared plus q over 2 quantity squared so that's going to be q squared over 4 equals r squared and now we've just got some algebra to do to solve this thing for q so let's go ahead and do that so i'm going to maybe factor out a half here and if we factor a half out of this we have to square it that's going to give us a fourth now i'll maybe multiply both sides of the equation by four and so that'll give us r plus q times root 3 squared plus q squared equals 4 r squared maybe that's a little simpler to work with at the moment now let's go ahead and multiply this out so that's going to give us r squared plus 2 r times q times root 3 plus 3 q squared plus q squared equals 4 r squared so that's what we get from multiplying this thing out right here now notice these two guys add up to four q squared and then we can move the four r squared over to the left hand side of the equation and then we have a quadratic equation that we can use to solve for q so let's go ahead and do that so we have 4 q squared and then what's our q term so our q term is going to be this thing right here so plus 2 r times the square root of 3. times q like that and then we'll have minus 3 r squared and that's from moving this 4r squared over to the other side of the equation and now that's equal to 0. so we can use the quadratic formula to solve for q where those are our coefficients so let's go ahead and do that we'll get q equals so it'll be negative this term so we have minus 2 r times the square root of 3 plus the square root of this term squared that's going to give us 12 r squared minus 4 times this term times that term so that's going to be plus 48r squared and then all over twice this term so that's going to be all over eight okay nice so now let's see what we can do we can simplify this quite a bit so we'll have minus 2 r times the square root of 3 plus the square root of 60 r squared great and then all over 8. the next thing that i want to notice is that i can factor an r out of this and write this as 15 times 4. we want to do that because 4 is a perfect square so we'll write this as 4 r squared times 15. so that means here we'll have minus 2r times the square root of 3 plus 2 r times the square root of 15 all over eight but now we can simplify this quite a bit and we will see that in the end we'll have q equals r times the quantity square root of 15 minus the square root of 3 over 4. and so we have achieved our goal which was to rewrite q in terms of r the radius of the circle and that's a good place to stop

Channel: Michael Penn

Views: 143,239

Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn

Id: 4ulzbICdw5Q

Length: 9min 40sec (580 seconds)

Published: Fri Sep 18 2020