Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.7 National olympiad Prove it

Given the polygons $P$ and $Q$ as shown in the grid below, cut $P$ into two polygons $P_1$ and $P_2$ such that, when pasted together differently, they form $Q$ .

Figure (Asymptote source)

import graph; size(16cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-2.05,xmax=15.10,ymin=-1.87,ymax=9.74; 
pen cqcqcq=rgb(0.75,0.75,0.75), zzttqq=rgb(0.6,0.2,0); 
draw((7,5)--(12,5)--(12,2)--(7,2)--cycle,zzttqq); draw((2,2)--(2,5)--(3,6)--(6,6)--(6,3)--(5,2)--cycle,zzttqq); 
/*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1;
for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); 
 draw((0,8)--(0,0)); draw((0,0)--(13,0)); draw((13,0)--(13,8)); draw((13,8)--(0,8)); draw((7,5)--(12,5),zzttqq); draw((12,5)--(12,2),zzttqq); draw((12,2)--(7,2),zzttqq); draw((7,2)--(7,5),zzttqq); draw((2,2)--(2,5),zzttqq); draw((2,5)--(3,6),zzttqq); draw((3,6)--(6,6),zzttqq); draw((6,6)--(6,3),zzttqq); draw((6,3)--(5,2),zzttqq); draw((5,2)--(2,2),zzttqq); 
dot((0,0),linewidth(1pt)+ds); dot((13,0),linewidth(1pt)+ds); dot((0,8),linewidth(1pt)+ds); dot((2,2),linewidth(1pt)+ds); dot((6,6),linewidth(1pt)+ds); dot((13,8),linewidth(1pt)+ds); dot((7,2),linewidth(1pt)+ds); dot((7,5),linewidth(1pt)+ds); dot((12,2),linewidth(1pt)+ds); dot((12,5),linewidth(1pt)+ds); label("$Q$",(8.42,2.56),NE*lsf,zzttqq); dot((5,2),linewidth(1pt)+ds); dot((6,3),linewidth(1pt)+ds); dot((2,5),linewidth(1pt)+ds); dot((3,6),linewidth(1pt)+ds); label("$P$",(4.65,2.74),NE*lsf,zzttqq); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);

Solution

1. **Identify the shapes of polygons $P$ and $Q$ :**
- Polygon $P$ is a hexagon with vertices at $(2,2)$ , $(2,5)$ , $(3,6)$ , $(6,6)$ , $(6,3)$ , and $(5,2)$ .
- Polygon $Q$ is a rectangle with vertices at $(7,2)$ , $(12,2)$ , $(12,5)$ , and $(7,5)$ .

2. **Determine the area of polygons $P$ and $Q$ :**
- The area of $Q$ is straightforward to calculate since it is a rectangle:
$\text{Area of } Q = \text{length} \times \text{width} = (12 - 7) \times (5 - 2) = 5 \times 3 = 15 \text{ square units}$
- To find the area of $P$ , we can use the Shoelace Theorem (Gauss's area formula for polygons):
$\text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n-1} (x_i y_{i+1} - y_i x_{i+1}) + (x_n y_1 - y_n x_1) \right|$
For $P$ with vertices $(2,2)$ , $(2,5)$ , $(3,6)$ , $(6,6)$ , $(6,3)$ , and $(5,2)$ :
$\text{Area} = \frac{1}{2} \left| 2 \cdot 5 + 2 \cdot 6 + 3 \cdot 6 + 6 \cdot 3 + 6 \cdot 2 + 5 \cdot 2 - (2 \cdot 2 + 5 \cdot 3 + 6 \cdot 6 + 6 \cdot 6 + 3 \cdot 5 + 2 \cdot 2) \right|$
$= \frac{1}{2} \left| 10 + 12 + 18 + 18 + 12 + 10 - (4 + 15 + 36 + 36 + 15 + 4) \right|$
$= \frac{1}{2} \left| 80 - 110 \right| = \frac{1}{2} \times 30 = 15 \text{ square units}$

3. **Cut polygon $P$ into two polygons $P_1$ and $P_2$ :**
- We need to find a way to cut $P$ such that the resulting pieces can be rearranged to form $Q$ .
- One possible way is to cut along the line from $(2,5)$ to $(6,3)$ . This will divide $P$ into two polygons:
- $P_1$ with vertices $(2,2)$ , $(2,5)$ , $(6,3)$ , and $(5,2)$ .
- $P_2$ with vertices $(2,5)$ , $(3,6)$ , $(6,6)$ , and $(6,3)$ .

4. **Rearrange $P_1$ and $P_2$ to form $Q$ :**
- Translate $P_1$ and $P_2$ such that they fit together to form the rectangle $Q$ .
- $P_1$ can be rotated and translated to fit the bottom part of $Q$ .
- $P_2$ can be rotated and translated to fit the top part of $Q$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.