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2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 8


Problem

[asy] pair C=(0,0),A=(0,12),B=(5,0); draw(C--A--B--C,dot); MP("A",A,N);MP("C",C,SW);MP("B",B,E); real r=10/3; pair X=(0,r),Y=X+r*(12/13,5/13); draw(X--Y,dot); MP("X",X,W);MP("Y",Y,NE); draw(circle(X,r)); MP("12",(0,8),W);MP("5",(2.5,0),S); [/asy]

Let $ABC$ be a right triangle with right angle at $C$. Suppose $AC = 12$ and $BC = 5$ and $CX$ is the diameter of a semicircle, where $X$ lies on $AC$ and the semicircle is tangent to side $AB$. Find the radius of the semicircle.

Solution 1

We can compute the area in two ways: $\triangle{ABC} = \frac{1}{2}\cdot 5\cdot 12 = 30$ or $\triangle{ABC}=\triangle{ABX} +\triangle{CBX} \frac{1}{2}\cdot{r}\cdot{13}+\frac{1}{2}\cdot{r}\cdot{5} = 9r$. Setting the two areas equal we obtain $r = \frac{10}{3}$.

Solution 2

Place the point C on the origin of the xy plane, $B$ at $(5,0)$ and $A$ at $(12,0)$. Point $X$ lies at cartesian coordinate $(0,r)$. The line AB has formula $y=\frac{12}{5}\cdot(5-x)$. The vector $\vec{XY}$ has coordinates $r(12/13,5/13)$ since it has length $r$ in the unit direction $(12,5)/13$ which is orthogonal to the line AB.

Then point Y has coordinates $X + \vec{XY}=(0,r)+r(12/13,5/13)$ and lies on the line $y=\frac{12}{5}\cdot(5-x)$. Substituting for these equations gives $r = \frac{10}{3}$.

Solution 3

Let $Z$ be the intersection of $\odot X$ and $AC$ that isn't $C$.

By the pythagorean theorem, AB = 13. By Power of a Point, $BC^2=BY^2$, so $5 = BC = BY$. Then, $AY=AB-BY=13-5=8$. Using PoP on $A$, we get $AY^2=AX\cdot AZ$. Simplifying, we get $AY^2=64=AX\cdot AZ=AZ\cdot 12$, so $AZ=\frac{16}{3}$. Since $r=\frac{AC-AZ}{2}$ and $AC=12$, $r=\frac{12-\frac{16}{3}}{2}=\frac{20}{6}=\boxed{\frac{10}{3}}$

See also

2017 UNM-PNM Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNM-PNM Problems and Solutions
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