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Difference between revisions of "2016 AMC 8 Problems/Problem 25"

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Let's call the triangle <math>\triangle ABC,</math> where <math>AB=16</math> and <math> AC=BC.</math> Let's say that <math>D</math> is the midpoint of <math>AB</math> and <math>E</math> is the point where <math>AC</math> is tangent to the semicircle. We could also use <math>BC</math> instead of <math>AC</math> because of symmetry.
 
Let's call the triangle <math>\triangle ABC,</math> where <math>AB=16</math> and <math> AC=BC.</math> Let's say that <math>D</math> is the midpoint of <math>AB</math> and <math>E</math> is the point where <math>AC</math> is tangent to the semicircle. We could also use <math>BC</math> instead of <math>AC</math> because of symmetry.
  
−
We notice that <math>\triangle ACD \cong \triangle BCD,</math> and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by <math>AA</math> similarity, <math>\triangle AED \sim \triangle ADC,</math> with <math>\angle EAD \cong \angle DAC</math> and <math> \angle CDA \cong \angle DEA.</math> This similarity means that we can create a proportion: <math>\frac{AD}{AB}=\frac{DE}{CD}.</math> We plug in <math>AD=\frac{AB}{2}=8, AC=17,</math> and <math>CD=15.</math> After we multiply both sides by <math>15,</math> we get <math>DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.</math>
+
Notice that <math>\triangle ACD \cong \triangle BCD,</math> and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by <math>AA</math> similarity, <math>\triangle AED \sim \triangle ADC,</math> with <math>\angle EAD \cong \angle DAC</math> and <math> \angle CDA \cong \angle DEA.</math> This similarity means that we can create a proportion: <math>\frac{AD}{AB}=\frac{DE}{CD}.</math> We plug in <math>AD=\frac{AB}{2}=8, AC=17,</math> and <math>CD=15.</math> After we multiply both sides by <math>15,</math> we get <math>DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.</math>
  
 
(By the way, we could also use <math>\triangle DEC \sim \triangle ADC.</math>)
 
(By the way, we could also use <math>\triangle DEC \sim \triangle ADC.</math>)

Revision as of 22:38, 14 December 2016

A semicircle is inscribed in an isosceles triangle with base $16$ and height $15$ so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?

[asy]draw((0,0)--(8,15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,180));[/asy]

$\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}$


Solution 1

Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height $15$ and base $\frac{16}{2} = 8$. The Pythagorean triple $8$-$15$-$17$ tells us that these triangles have hypotenuses of $17$.

Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be $r$.

The area of the entire isosceles triangle is $\frac{(16)(15)}{2} = 120$, so the area of each of the two congruent right triangles it gets split into is $\frac{120}{2} = 60$. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is $\frac{17r}{2}$. Thus we can write the equation $\frac{17r}{2} = 60$, so $17r = 120$, so $r = \boxed{\textbf{(B) }\frac{120}{17}}$.


Solution 2

First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, $60$. $\frac{60}{17}$ times $2$ get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is $\boxed{\textbf{(B) }\frac{120}{17}}$.


Solution 3: Similar Triangles

[asy] pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NW);[/asy] Let's call the triangle $\triangle ABC,$ where $AB=16$ and $AC=BC.$ Let's say that $D$ is the midpoint of $AB$ and $E$ is the point where $AC$ is tangent to the semicircle. We could also use $BC$ instead of $AC$ because of symmetry.

Notice that $\triangle ACD \cong \triangle BCD,$ and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by $AA$ similarity, $\triangle AED \sim \triangle ADC,$ with $\angle EAD \cong \angle DAC$ and $\angle CDA \cong \angle DEA.$ This similarity means that we can create a proportion: $\frac{AD}{AB}=\frac{DE}{CD}.$ We plug in $AD=\frac{AB}{2}=8, AC=17,$ and $CD=15.$ After we multiply both sides by $15,$ we get $DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.$

(By the way, we could also use $\triangle DEC \sim \triangle ADC.$)

Solution 4: Inscribed Circle

[asy] pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label("$B$",B,SW); label("$D$",D,SE); label("$A$",A,N); label("$M$",M,S); label("$C$",C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));[/asy]

We'll call this triangle $\triangle ABD$. Let the midpoint of base $BD$ be $M$. Divide the triangle in half by drawing a line from $A$ to $M$. Half the base of $\triangle ABD$ is $\frac{16}{2} = 8$. The height is $15$, which is given in the question. Using the Pythagorean Triple $8$-$15$-$17$, the length of each of the legs ($AB$ and $DA$) is 17.

Reflect the triangle over its base. This will create an inscribed circle in a rhombus $ABCD$. Because $AB \cong DA$, $BC \cong CD$. Therefore $AB = BC = CD = DA$.

The semiperimeter $s$ of the rhombus is $\frac{AB + BC + CD + DA}{2} = \frac{(17)(4)}{2} = 34$. Since the area of $\triangle ABD$ is $\frac{bh}{2}$, the area $[ABCD]$ of the rhombus is twice that, which is $bh = (16)(15) = 240$.

The Formula for the Incircle of a Quadrilateral is $s$$r$ = $[ABCD]$. Substituting the semiperimeter and area into the equation, $34r = 240$. Solving this, $r = \frac{240}{34}$ = $\boxed{\textbf{(B) }\frac{120}{17}}$.

Part Solution 5: Answer Choices

Part Solution by e_power_pi_times_i

Notice that the radius must be smaller than half the base (which is $8$). Therefore answer choices $C$, $D$, and $E$ are eliminated. If you guess at this point, there is a $\dfrac{1}{2}$ chance that you will get it right.

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