CEMC Pascal, 2015

In each diagram, we label the origin $(0,0)$ as $O$, the point $(4,0)$ as $A$, the point $(4,4)$ as $B$, and the point $(0,4)$ as $C$.

Thus, in each diagram, square $OABC$ is 4 by 4 and so has area $16$.

In the first diagram, we label $(1,4)$ as $E$ and $(4,1)$ as $F$.

In the second diagram, we label $(0,1)$ as $G$ and $(3,0)$ as $H$.

In the third diagram, we label $(2,0)$ as $J$ and $(4,3)$ as $K$.

In the first diagram, the area of $\triangle OEF$ equals the area of square $OABC$ minus the areas of $\triangle OCE$, $\triangle EBF$ and $\triangle FAO$.

Each of these three triangles is right-angled at a corner of the square.

Since $OC=4$ and $CE=1$, the area of $\triangle OCE$ is $\frac{1}{2}(4)(1)=2$.

Since $EB=3$ and $BF=3$, the area of $\triangle EBF$ is $\frac{1}{2}(3)(3)=\frac{9}{2}$.

Since $FA=1$ and $AO=4$, the area of $\triangle FAO$ is $\frac{1}{2}(1)(4)=2$.

Therefore, the area of $\triangle OEF$ equals $16-2-\frac{9}{2}-2=\frac{15}{2}$, or $m = \frac{15}{2}$.

In the second diagram, the area of $\triangle GBH$ equals the area of square $OABC$ minus the areas of $\triangle GCB$, $\triangle BAH$ and $\triangle HOG$.

Each of these three triangles is right-angled at a corner of the square.

Since $GC=3$ and $CB=4$, the area of $\triangle GCB$ is $\frac{1}{2}(3)(4)=6$.

Since $BA=4$ and $AH=1$, the area of $\triangle BAH$ is $\frac{1}{2}(4)(1)=2$.

Since $HO=3$ and $OG=1$, the area of $\triangle HOG$ is $\frac{1}{2}(1)(3)=\frac{3}{2}$.

Therefore, the area of $\triangle HOG$ equals $16-6-2-\frac{3}{2}=\frac{13}{2}$, or $n = \frac{13}{2}$.

In the third diagram, the area of $\triangle CKJ$ equals the area of square $OABC$ minus the areas of $\triangle CBK$, $\triangle KAJ$ and $\triangle JOC$.

Each of these three triangles is right-angled at a corner of the square.

Since $CB=4$ and $BK=1$, the area of $\triangle CBK$ is $\frac{1}{2}(4)(1)=2$.

Since $KA=3$ and $AJ=2$, the area of $\triangle KAJ$ is $\frac{1}{2}(3)(2)=3$.

Since $JO=2$ and $OC=4$, the area of $\triangle JOC$ is $\frac{1}{2}(2)(4)=4$.

Therefore, the area of $\triangle CKJ$ equals $16-2-3-4=7$, or $p = 7$.

Since $m=\frac{15}{2}=7\frac{1}{2}$, and $n = \frac{13}{2}=6\frac{1}{2}$, and $p=7$, then $\text{n lt;p lt;m}$.