Question 248·Hard·Math

Let $a=8x-22$ be the measure of each acute angle, and let $b$ be the measure of an obtuse angle. At either crossing, an acute angle and an obtuse angle beside it form a linear pair: together they make a straight angle. So $a+b=180$.

The problem counts four acute angles and one obtuse angle. Using $a$ and $b$ for their measures, that total is $4a+b$. It is also given as $24x+w$, so

Break the four copies of $a$ into three copies and one more. That puts one acute angle beside the obtuse angle in the equation:

Since $a+b=180$, trade the acute-obtuse pair for $180$: $3a+180=24x+w$. The $180$ replaces both angles in the pair, not the obtuse angle alone.

Substitute $a=8x-22$ into the total: $3(8x-22)+180=24x+w$. This equality must hold for every valid $x$, because both sides describe the same five-angle sum.

Type $x_1=[3...7]$, then $3(8x_1-22)+180\sim24x_1+w$. The list gives Desmos several inputs where $8x_1-22$ is acute; $\sim$ asks it to fit the constant $w$ so the totals match. Under PARAMETERS, Desmos shows $w=114$. So the value of $w$ is $114$. Grid in 114.

Instead of fitting $w$, compare the parts of $3(8x-22)+180$ with $24x+w$. Distribute $3$: $24x-66+180=24x+w$. The $24x$ terms already match.

Subtract $24x$ from both sides: $-66+180=w$. So $w$ comes from the constant parts of the acute angles and the straight angle.

Add the constants: $114=w$. So $w=114$. Grid in 114.