, 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 $b<2$, 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 $AD>0$, 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 $CD > 0$, 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 $\frac{1}{2}(DB+OA)(AB) = \frac{1}{2}(3+6)(6)=27$.