2013 MTAP Saturday Class 4th Year S-5 Challenge Problem Solution

When I was coaching my Alma-matter’s team for MMC last year. A teacher gave me this paper. I realize to make a solution and publish it here. This could be a help for people doing a self-review.

MTAP has nothing to do with this solution. This is only MY SOLUTION. Anyone who can read this is allowed to make violent reactions/clarifications.

Problem 1.

fig1The figure is a circumscribed polygon WXYZ

where XY=8, YZ=10, WZ=21.

Find WX.

 

 

Solution:

Label the points of intersections of the circle to the quadrilateral like the figure shown below. Let x=CX and y=CW. XW=x+y.

 

fig2

From the figure, CX=DX, DY=EY, EZ=FZ, FW=CW.

Why? Tangents of the same circle and the same external point are congruent.

Since CX=x, DX=x, DY=8-x.
Similarly, CW=y, FW=y, FZ=21-y
Since DY=EY, EY=8-x
Since FZ=EZ, EZ=21-y
But EY+EZ=10

EY+EZ=10
8-x+21-y=10
29-10=x+y
\boxed{x+y=19}

Since XW=x+y, therefore XW=19.

Problem 2:

fig3Problem 2: XU and XV are tangents to circle O. XO=17, OU=8

Find XZ+YZ+XY.

Solution:

fig4

Let’s connect X to O. The line must pass through W as shown in the figure above.

Observe that \triangle XUO\sim\triangle XWZ.\angle{XUO}=\angle{XWZ}=90º by definition of the tangent line and radius.

From \triangle XUO, we can solve for XU

XO^2=XU^2+UO^2
17^2=XU^2+8^2
XU=15

Observe that WO is also radius of the circle, thus WO=8.

XO=OW+XW
17=8+XW
XW=9

By similar triangle, we can solve for XZ and WZ.
Solving for XZ

\displaystyle\frac{XZ}{XW}=\displaystyle\frac{XO}{XU}
\displaystyle\frac{XZ}{9}=\displaystyle\frac{17}{15}
XZ=\displaystyle\frac{17}{15}\cdot 9
XZ=\displaystyle\frac{51}{5}

Solving for YZ

YZ=2\cdot WZ
YZ=2\cdot \displaystyle\frac{24}{5}
YZ=\displaystyle\frac{48}{5}

Solving for XY

XY=XZ=\displaystyle\frac{51}{5}
Thus,

XZ+YZ+XY=\displaystyle\frac{51}{5}+\displaystyle\frac{48}{5}+\displaystyle\frac{51}{5}
\boxed{XZ+YZ+XY=30}

Problem 3:
fig5An equilateral triangle with sides of length 10 cm is inscribed in a circle. Find the distance from the center of the circle to the side of triangle.

 

 

Solution:

The distance that asked here is the perpendicular or the shortest distance from the center of the circle to the side of the triangle.

fig6Draw and label the triangle as shown. Since the given triangle is equilateral the perpendicular distance from the center to the side of the triangle contains the height on the triangle.

In the figure, \triangle GOB forms a 30-60-90 triangle. From here, we know that

GB=OG\sqrt{3}

5=x\sqrt{3}

x=\displaystyle\frac{5}{\sqrt{3}}

\boxed{x=\displaystyle\frac{5\sqrt{3}}{3}}

Problems 4-10 will be solved in the following days. Subscribe to our newsletter by entering your email address in upper right side of this page.

 

 

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