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Rotation: Difference between revisions

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Created page with "A rotation of a planar figure is a transformation that preserves area and angles, but not orientation. The resulting figure is congruent to the first. Suppose we wish to rotate..."
 
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We would first draw segment <math>AO</math>. Then, we would draw a new segment, <math>A'O</math> such that the angle formed is <math>60^{\circ}</math>, and <math>AO=A'O</math>. Do this for points <math>B</math> and <math>C</math>, to get the new triangle <math>A'B'C'</math>
We would first draw segment <math>AO</math>. Then, we would draw a new segment, <math>A'O</math> such that the angle formed is <math>60^{\circ}</math>, and <math>AO=A'O</math>. Do this for points <math>B</math> and <math>C</math>, to get the new triangle <math>A'B'C'</math>


{{stub}}
=== Practice Problems ===
*Isosceles <math>\triangle ABC</math> has a right angle at <math>C</math>.  Point <math>P</math> is inside <math>\triangle ABC</math>, such that <math>PA=11</math>, <math>PB=7</math>, and <math>PC=6</math>. Legs <math>\overline{AC}</math> and <math>\overline{BC}</math> have length <math>s=\sqrt{a+b\sqrt{2}}</math>, where <math>a</math> and <math>b</math> are positive integers.  What is <math>a+b</math>?
 
<asy>
pathpen = linewidth(0.7);
pen f = fontsize(10);
size(5cm);
pair B = (0,sqrt(85+42*sqrt(2)));
pair A = (B.y,0);
pair C = (0,0);
pair P = IP(arc(B,7,180,360),arc(C,6,0,90));
D(A--B--C--cycle);
D(P--A);
D(P--B);
D(P--C);
MP("A",D(A),plain.E,f);
MP("B",D(B),plain.N,f);
MP("C",D(C),plain.SW,f);
MP("P",D(P),plain.NE,f);
</asy>
 
<math>
\mathrm{(A)}\ 85
\qquad
\mathrm{(B)}\ 91
\qquad
\mathrm{(C)}\ 108
\qquad
\mathrm{(D)}\ 121
\qquad
\mathrm{(E)}\ 127
</math>
 
([[2006 AMC 12B Problems/Problem 23|Source]])

Revision as of 22:17, 13 January 2021

A rotation of a planar figure is a transformation that preserves area and angles, but not orientation. The resulting figure is congruent to the first.

Suppose we wish to rotate triangle $ABC$ $60^{\circ}$ clockwise around a point $O$, also known as the center of rotation.

We would first draw segment $AO$. Then, we would draw a new segment, $A'O$ such that the angle formed is $60^{\circ}$, and $AO=A'O$. Do this for points $B$ and $C$, to get the new triangle $A'B'C'$

Practice Problems

  • Isosceles $\triangle ABC$ has a right angle at $C$. Point $P$ is inside $\triangle ABC$, such that $PA=11$, $PB=7$, and $PC=6$. Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$, where $a$ and $b$ are positive integers. What is $a+b$?

[asy] pathpen = linewidth(0.7); pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f); [/asy]

$\mathrm{(A)}\ 85 \qquad \mathrm{(B)}\ 91 \qquad \mathrm{(C)}\ 108 \qquad \mathrm{(D)}\ 121 \qquad \mathrm{(E)}\ 127$

(Source)

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