Question 2·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

A tangent touches the circle at one point. Tangent segments from the same outside point are equal, so $TA=TB$. Also, $CA=CB$ because both are radii from the center to the circle. Both $C$ and $T$ lie on the perpendicular bisector of $AB$. That line crosses $AB$ at its midpoint and at a right angle. Call the midpoint $M$; then $C$, $M$, and $T$ line up, and $CM\perp AB$.

Since $M$ is the midpoint of the $40$-centimeter chord, $AM=20$. Triangle $CMA$ is right at $M$, and its hypotenuse $CA$ is the $25$-centimeter radius. The 15-20-25 right-triangle pattern gives $CM=15$ centimeters.

The radius $CA$ is perpendicular to tangent $AT$ at $A$, so $\angle CAT$ is right, like $\angle CMA$. The triangles also share angle $C$ because $C$, $M$, and $T$ line up. By angle-angle similarity, two matching angles make the triangles the same shape: $\triangle CMA\sim\triangle CAT$. The vertices match $C$ with $C$, $M$ with $A$, and $A$ with $T$.

In those similar triangles, $CM$ matches $CA$, and $MA$ matches $AT$. Matching sides keep the same ratio, so:

Substitute the lengths you found:

Let $x_1$ be the positive length $AT$. Type $\frac{15}{25}\sim\frac{20}{x_1}\{x_1>0\}$ in Desmos: $\sim$ tells it to fit the unknown, and the restriction keeps a length positive. Desmos shows $x_1\approx33.3333$ under PARAMETERS. Type $x_1$ on the next line and tap the fraction button to see $\frac{100}{3}$. So $AT=\frac{100}{3}$ centimeters, not the $20$-centimeter half-chord.

Quadrilateral $CATB$ has two $25$-centimeter radii and two equal tangent segments. The chord $AB$ crosses its interior, so it isn’t a perimeter side. Type $2(25)+2x_1$ on the next Desmos line. It shows $116.666\ldots$; tap the fraction button for $\frac{350}{3}$. The perimeter is $\frac{350}{3}$ centimeters. Grid in 350/3.