What is the sum of the radii of the three incircles
A 2 0 × 8 rectangle is partitioned into three right triangles, and a circle is inscribed in each triangle.
Find the sum of the radii of these 3 circles.
The answer is 8.
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3 solutions
I guessed correctly because I figured the 3 circles would amount to the same space as 2 circles filling the rectangle. It was purely intuitive and lucky.
@Jayson Powers - Steven's proof of the inradius of a right triangle is correct, but we can show it without resorting to the Pythagorean Theorem. Suppose that the incircle divides the hypotenuse into lengths p and q, so that p+q=c. Then by similar triangles we have p+r=a and q+r=b. Adding these and subtracting the first equation gives the result.
I'm curious if anyone knows a proof for inradius = (a+b-c)/2?
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There's many ways to prove that the inradius of a right triangle is 2 a + b − c . Here's one involving area:
Let a right triangle have legs a , b and hypotenuse c , and let r be the inradius of this triangle. By the triangle area formulae, the area of a right triangle is equal to both 2 a b and r s , where s = 2 a + b + c .
Now, let's take the Pythagorean Theorem and manipulate it to get an expression for a b in terms of s :
a 2 + b 2 a 2 + 2 a b + b 2 ( a + b ) 2 ( a + b ) 2 − c 2 ( a + b + c ) ( a + b − c ) a b a b = c 2 = c 2 + 2 a b = c 2 + 2 a b = 2 a b = 2 a b = 2 ( a + b + c ) ( a + b − c ) = s ( a + b − c ) .
Thus,
2 a b 2 s ( a + b − c ) r = r s = r s = 2 a + b − c . ■
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And where does this A=rs come from? I've never heard of inradius before.
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@Paul Blohn – Draw a triangle with sides a , b , c . Draw its angle bisectors to find the incenter. Now draw the three inradii of length r . You ought to see that the triangle's area is the sum of three inner triangular areas r a / 2 , r b / 2 , and r c / 2 . Or, A = r 2 a + b + c = r s .
@Paul Blohn – Check out this wiki page for info. The inradius is the radius of the incircle of a triangle, which is the circle inside the triangle that is tangent to all three sides of the triangle.
It's quite easy to solve the question once you know the inradius formula...
On your generalization note: is this a definition or is this a theorem?
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This is a theorem. You can prove it by using the same steps in the solution. The condition x ≥ 2 y is there because if it weren't, then there wouldn't be a point X on C D such that ∠ A X B is a right angle.
Just one issue here. As far as I can see only the illustration actually limits the circles to be incircles (by the way that concept is not a worldwide concept). They can just as well be inscribed outcircles (yep I am aware that the term inscribed becomes a bit weird there)
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Problem states that "a circle is inscribed in each triangle". When you say "inscribed outcircles" do you mean circumcircles or what? Circumcircles aren't inscribed - they're circumscribed.
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no I mean inscribed circles that in most peoples minds is outside the triangle. I have noticed that in English you have incircles and outcircles and circumcircles and the outcircles are inscribed circles from their properties (all triangle sides are tangent to each of them).
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@Steen Grode – Are you talking about excircles ? Those are the only circles, other than the incircle, that are tangent to all three sides of a triangle, if we extend the sides as far as we need to. Unfortunately, they aren't inscribed because they don't exist in the interior of any triangle.
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@Steven Yuan – Strictly speaking Euclide you can define a specific inside of a triangle in 2D and by that you can distiguish between the four "tangent" circles to any triangle. Apart from that case it is not possible and you have to define this first as it is not a general definition. In the question I can just as well take what most people think as an inside as an outside. That was why I said there is nothing but the drawing that constrains the question to just one solution.
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@Steen Grode – I don't quite follow your logic. Inscribed circles are almost a universally understood idea. Even then, one part of the definition of an inscribed figure is that it is entirely contained within a closed shape that existed before we constructed the figure. So, we can't say that an excircle is inscribed because it's not inside any pre-constructed closed shape.
I would like to express this concern that the figure is quite misleading As I didn't know the incircle formula I tried to combine trig. And Algebra And I found that all the triangles were isoceles ones i.e, all the angles in the triangle were 45°s by trignometric ratios. This implies that the figure given is wrong Please give a counter intuitive proof, if any ,to show that the angles are no that way.
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If all the acute angles were 45 degrees, X would be the midpoint of DC, which it clearly isn't.
Can you show me the work you did to reach your conclusion?
Sorry , I reviewed my work and noticed a mistake in one of the equations and it is clearly not 45°s Thank u
'A 20x8 rectangle is partitioned into three right triangles.'
It would help to add the words 'as shown' at the end of the sentence above, because the provided sketch is not the only way to do such a partitioning. Instead, you could draw in a full diagonal of the rectangle, splitting it into two identical right triangles. Then further partition one of those right triangles by dropping down an altitude from its right angle.
The three right triangles that result from this new partitioning are not congruent to the three shown above. (The key ratios are 2:5:sqrt(29) in the second partitioning compared to 1:2:sqrt(5) in the original partitioning.) Following logic identical to what Steven Yuan laid out, the sum of the three radii comes out to be 40sqrt(29)/29 in the new case. [Generalization: x*y/sqrt(x^2 + y^2) = area/diagonal.]
This new partitioning has the added benefit of working for rectangles of any dimensions, whereas the partitioning shown above only works for long, thin rectangles (x >= 2y), as Steven pointed out.
The problem with the language "as shown" is that it can be taken to mean the figure is drawn to scale.
Let's suppose you draw a diagonal to create a triangle using one half of the rectangle. And then you draw an arbitrary segment from the opposite corner to the diagonal, creating two triangles. Clearly the sum of the radii will depend on the point chosen. So clearly this line of argument leads to a non-unique solution.
Or you could just draw basic inferences from the figure.
Relevant wiki: Pythagorean Theorem
Using the Pythagorean Theorem for three right triangles we come to the result that the vertex of the biggest triangle has divided the side of the rectangle by two pieces with the lengths of 4 and 16. Thus the triangles from right to left have the sides of 8, 4, 4√5 ; 4√5 , 8√5 , 20, and 16, 8, 8√5 respectively. Using formula r =A/p, where r is the radius of a circle inscribed in the triangle, A is the area of the triangle, and p is half of the perimeter of the triangle, we find the three radii of the circles: 8/(3+√5), 40/(5+3√5), 16/(3+√5) . The sum of all of them is 8 (good exercise in algebra!).
Obviously this solution makes sense to use if the question in the problem would be: “Find the radius of each inscribed circle”.
It'd be helpful if you attached a diagram
Relevant wiki: Incircle of Triangle
Label the points as shown in the diagram.
The inradius of a right triangle with legs a , b and hypotenuse c is equal to 2 a + b − c . Thus, the sum of the inradii of △ A D X , △ B C X , and △ A B X equals
2 A D + D X − A X + 2 B C + C X − B X + 2 A X + B X − A B = 2 8 + D X − A X + 2 8 + ( 2 0 − D X ) − B X + 2 A X + B X − 2 0 = 2 8 + 8 + 2 0 − 2 0 = 8 .
Generalization : If A B = x , B C = y , with x ≥ 2 y , then the sum of the inradii of the three triangles is equal to y .