Problem - 03 - National Math Camp 2021 Combinatorics Test - "All you see is isosceles"
Posted: Fri Apr 30, 2021 5:25 pm
Show that for all positive integers $n\geq8$, you can cut any quadrilateral into $n$ isosceles triangles.
First, let me give a claim,
Claim-1:
For all $n\geq 4$ we can cut a triangle into $n$ isosceles triangles.
Proof:
Base Case:
Let us take the largest angle of a given triangle and drop an altitude on the opposite side. The altitude must be inside the triangle. So our initial triangle has become $2$ right-angled triangles. Which we can cut into $4$ isosceles triangles by cutting the line from the circumcentre of each right-angled triangle and its vertices. So our base case is done.
Inductive hypothesis:
Let we can cut any given triangle in $k$ isosceles triangle.
Inductive step:
Let, $\triangle ABC$ be a triangle and where $AC$ is its longest side. Now we take $A$ or $C$ as the centre and $AB$(If we took $A$) or $BC$(If we took $B$) as radius and draw a circle that intersects $AC$ at $X$. Now we can cut through $BC$ and get an isosceles triangle $AXB$ if we chose $A$ or $CBX$ if we chose $C$. Depending on our choice one of them is an isosceles triangle and the other is another triangle which we can cut into $k$ isosceles triangle[By our inductive hypothesis]. So we can cut $\triangle ABC$ into $k+1$ isosceles triangles. $\square$
Now the rest is easy. We just draw a diagonal of the quadrilateral and have 2 triangles which we can cut into any number of isosceles triangles $k\geq 4$ by claim-1. So we can cut any quadrilateral in $n \geq 8$ triangles. $\square$
Claim-1:
For all $n\geq 4$ we can cut a triangle into $n$ isosceles triangles.
Proof:
Base Case:
Let us take the largest angle of a given triangle and drop an altitude on the opposite side. The altitude must be inside the triangle. So our initial triangle has become $2$ right-angled triangles. Which we can cut into $4$ isosceles triangles by cutting the line from the circumcentre of each right-angled triangle and its vertices. So our base case is done.
Inductive hypothesis:
Let we can cut any given triangle in $k$ isosceles triangle.
Inductive step:
Let, $\triangle ABC$ be a triangle and where $AC$ is its longest side. Now we take $A$ or $C$ as the centre and $AB$(If we took $A$) or $BC$(If we took $B$) as radius and draw a circle that intersects $AC$ at $X$. Now we can cut through $BC$ and get an isosceles triangle $AXB$ if we chose $A$ or $CBX$ if we chose $C$. Depending on our choice one of them is an isosceles triangle and the other is another triangle which we can cut into $k$ isosceles triangle[By our inductive hypothesis]. So we can cut $\triangle ABC$ into $k+1$ isosceles triangles. $\square$
Now the rest is easy. We just draw a diagonal of the quadrilateral and have 2 triangles which we can cut into any number of isosceles triangles $k\geq 4$ by claim-1. So we can cut any quadrilateral in $n \geq 8$ triangles. $\square$