, 2018

Suppose that the five-digit integer has digits $abcde$.

The digits $a,b,c,d,e$ are $1,3,5,7,9$ in some order.

Since $abcde$ is greater than 80 000, then $a\geq 8$, which means that $a=9$.

Since $9bcde$ is less than $92\,000$, then $\text{b lt;2}$, which means that $b=1$.

Since $91cde$ has units (ones) digit 3, then $e=3$.

So far, the integer is $91cd3$, which means that $c$ and $d$ are $5$ and $7$ in some order.

Since the two-digit integer $cd$ is divisible by 5, then it must be $75$.

This means that the the five-digit integer is $91753$.
By the Pythagorean Theorem in $\triangle ADB$, $$AD^2 = AB^2 - BD^2 = 13^2 - 12^2 = 169-144 = 25$$ Since $\text{AD gt;0}$, then $AD = \sqrt{25}=5$.

By the Pythagorean Theorem in $\triangle CDB$, $$CD^2 = BC^2 - BD^2 = (12\sqrt{2})^2 - 12^2 = 12^2(2) - 12^2 = 12^2$$ Since $\text{CD gt; 0}$, then $CD=12$.

Therefore, $AC=AD+DC=5+12=17$.
Solution 1

The area of the shaded region equals the area of square $OABC$ minus the area of $\triangle OCD$.

Since square $OABC$ has side length 6, then its area is $6^2$ or $36$.

Also, $OC=6$.

Since the equation of the line is $y=2x$, then its slope is $2$.

Since the slope of the line is $2$, then $\dfrac{OC}{CD}=2$.

Since $OC=6$, then $CD=3$.

Thus, the area of $\triangle OCD$ is $\frac{1}{2}(OC)(CD) = \frac{1}{2}(6)(3)=9$.

Finally, the area of shaded region must be $36 - 9 = 27$.

Solution 2

Since square $OABC$ has side length $6$, then $OA=AB=CB=OC=6$.

Since the slope of the line is $2$, then $\dfrac{OC}{CD}=2$.

Since $OC=6$, then $CD=3$.

Since $CB=6$ and $CD=3$, then $DB=CB-CD=3$.

The shaded region is a trapezoid with parallel sides $DB=3$ and $OA=6$ and height $AB=6$.

Therefore, the area of the shaded region is $\frac{1}{2}(DB+OA)(AB) = \frac{1}{2}(3+6)(6)=27$.