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Difference between revisions of "1992 AHSME Problems/Problem 25"

(Created page with "== Problem == In <math>\triangle{ABC}</math>, <math>\angle{ABC=120^\circ,AB=3</math> and <math>BC=4</math>. If perpendiculars constructed to <math>\overline{AB}</math> at <math>...")
 
m (→‎Solution 2 (Cyclic Quadrilaterals, Right Triangles, LoC))
 
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== Problem ==
 
== Problem ==
  
−
In <math>\triangle{ABC}</math>, <math>\angle{ABC=120^\circ,AB=3</math> and <math>BC=4</math>. If perpendiculars constructed to <math>\overline{AB}</math> at <math>A</math> and to <math>\overline{BC}</math> at <math>C</math> meet at <math>D</math>, then <math>CD=</math>
+
In <math>\triangle{ABC}</math>, <math>\angle ABC=120^\circ,AB=3</math> and <math>BC=4</math>. If perpendiculars constructed to <math>\overline{AB}</math> at <math>A</math> and to <math>\overline{BC}</math> at <math>C</math> meet at <math>D</math>, then <math>CD=</math>
  
 
<math>\text{(A) } 3\quad
 
<math>\text{(A) } 3\quad
Line 9: Line 9:
 
\text{(E) } \frac{10}{\sqrt{3}}</math>
 
\text{(E) } \frac{10}{\sqrt{3}}</math>
  
−
== Solution ==
+
== Solution 1 (Extending Line Segments) ==
−
<math>\fbox{E}</math>
+
We begin by drawing a diagram.
 +
<asy>
 +
import olympiad; import cse5; import geometry; size(150);
 +
defaultpen(fontsize(10pt));
 +
defaultpen(0.8);
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dotfactor = 4;
 +
pair A = origin;
 +
pair C = A+dir(55);
 +
pair D = A+dir(0);
 +
pair B = extension(A,A+dir(90),C,C+dir(-155));
 +
label("$A$",A,S);
 +
label("$C$",C,NE);
 +
label("$D$",D,SE);
 +
label("$B$",B,NW);
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label("$4$",B--C,NW);
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label("$3$",A--B,W);
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draw(A--C--D--cycle);
 +
draw(A--B--C);
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draw(rightanglemark(B,C,D,2));
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draw(rightanglemark(B,A,D,2));
 +
</asy>
 +
We extend <math>CB</math> and <math>DA</math> to meet at <math>E.</math> This gives us a couple right triangles in <math>CED</math> and <math>BEA.</math>
 +
<asy>
 +
import olympiad; import cse5; import geometry; size(250);
 +
defaultpen(fontsize(10pt));
 +
defaultpen(0.8);
 +
dotfactor = 4;
 +
pair A = origin;
 +
pair C = A+dir(55);
 +
pair D = A+dir(0);
 +
pair B = extension(A,A+dir(90),C,C+dir(-155));
 +
pair E = extension(A,A+2*dir(180),B,B+2*dir(-155));
 +
label("$A$",A,S);
 +
label("$C$",C,NE);
 +
label("$D$",D,SE);
 +
label("$B$",B,NW);
 +
label("$4$",B--C,NW);
 +
label("$3$",A--B,W);
 +
label("$E$",E,SW);
 +
draw(A--C--D--cycle);
 +
draw(A--B--C);
 +
draw(rightanglemark(B,C,D,2));
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draw(rightanglemark(B,A,D,2));
 +
draw(A--E--B,dashed);
 +
</asy>
 +
We see that <math>\angle E = 30^\circ</math>. Hence, <math>\triangle BEA</math> and <math>\triangle DEC</math> are 30-60-90 triangles.
 +
 
 +
Using the side ratios of 30-60-90 triangles, we have <math>BE=2BA=6</math>. This tells us that <math>CE=BC+BE=4+6=10</math>. Also, <math>EA=3\sqrt{3}</math>.
 +
 
 +
Because <math>\triangle DEC\sim\triangle BEA</math>, we have <cmath>\frac{10}{3\sqrt{3}}=\frac{CD}{3}.</cmath>
 +
Solving the equation, we have
 +
<cmath>\begin{align*}
 +
\frac{CD}3&=\frac{10}{3\sqrt{3}}\\
 +
CD&=3\cdot\frac{10}{3\sqrt{3}}\\
 +
CD&=\frac{10}{\sqrt{3}}\
 +
\end{align*}</cmath>
 +
Hence, <math>CD=\boxed{\textbf{E}}</math>.
 +
 
 +
== Solution 2 (Cyclic Quadrilaterals, Right Triangles, LoC) ==
 +
 
 +
Since <math>\angle{A}+\angle{C} = 180^{\circ}, ABCD</math> is cyclic. Using Ptolemy's Theorem gets <math>\text{(1): } 4AD + 3CD = \sqrt{37}BD.</math>
 +
 
 +
Right triangles <math>\Delta ABD</math> and <math>\Delta CBD</math> obtain <math>\text{(2): } 9+AD^2=BD^2</math> and <math>\text{(3):} 16+CD^2=BD^2,</math> respectively.
 +
 
 +
Seeing squares in <math>\text{(2)}</math> and <math>\text{(3)}</math>, we square <math>\text{(1)}</math>  and get <math>\text{(4): } 16AD^2+9CD^2 + 24AD\cdot CD = 37BD^2.</math>
 +
 
 +
We don't like that <math>AD\cdot CD</math> term, but fortunately LoC exists: <math>37 = AD^2 + CD^2 - 2*AD*CD*\cos(60^{\circ})</math>. Solving for <math>AD\cdot CD</math> and plugging it into <math>\text{(4)}</math>, and using <math>AD^2 = 7 + CD^2</math> and <math>BD^2 = 16 + CD^2</math> from the first two equations, gets <math>40(CD^2+7)+33CD^2 - 24\cdot 37 = 37(16+CD^2).</math>
 +
 
 +
Solve for <math>CD = \boxed{\textbf{E}}</math>.
 +
 
 +
~PureSwag
  
 
== See also ==
 
== See also ==

Latest revision as of 15:23, 24 October 2022

Problem

In $\triangle{ABC}$, $\angle ABC=120^\circ,AB=3$ and $BC=4$. If perpendiculars constructed to $\overline{AB}$ at $A$ and to $\overline{BC}$ at $C$ meet at $D$, then $CD=$

$\text{(A) } 3\quad \text{(B) } \frac{8}{\sqrt{3}}\quad \text{(C) } 5\quad \text{(D) } \frac{11}{2}\quad \text{(E) } \frac{10}{\sqrt{3}}$

Solution 1 (Extending Line Segments)

We begin by drawing a diagram. [asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = A+dir(55); pair D = A+dir(0); pair B = extension(A,A+dir(90),C,C+dir(-155)); label("$A$",A,S); label("$C$",C,NE); label("$D$",D,SE); label("$B$",B,NW); label("$4$",B--C,NW); label("$3$",A--B,W); draw(A--C--D--cycle); draw(A--B--C); draw(rightanglemark(B,C,D,2)); draw(rightanglemark(B,A,D,2)); [/asy] We extend $CB$ and $DA$ to meet at $E.$ This gives us a couple right triangles in $CED$ and $BEA.$ [asy] import olympiad; import cse5; import geometry; size(250); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = A+dir(55); pair D = A+dir(0); pair B = extension(A,A+dir(90),C,C+dir(-155)); pair E = extension(A,A+2*dir(180),B,B+2*dir(-155)); label("$A$",A,S); label("$C$",C,NE); label("$D$",D,SE); label("$B$",B,NW); label("$4$",B--C,NW); label("$3$",A--B,W); label("$E$",E,SW); draw(A--C--D--cycle); draw(A--B--C); draw(rightanglemark(B,C,D,2)); draw(rightanglemark(B,A,D,2)); draw(A--E--B,dashed); [/asy] We see that $\angle E = 30^\circ$. Hence, $\triangle BEA$ and $\triangle DEC$ are 30-60-90 triangles.

Using the side ratios of 30-60-90 triangles, we have $BE=2BA=6$. This tells us that $CE=BC+BE=4+6=10$. Also, $EA=3\sqrt{3}$.

Because $\triangle DEC\sim\triangle BEA$, we have \[\frac{10}{3\sqrt{3}}=\frac{CD}{3}.\] Solving the equation, we have \begin{align*} \frac{CD}3&=\frac{10}{3\sqrt{3}}\\ CD&=3\cdot\frac{10}{3\sqrt{3}}\\ CD&=\frac{10}{\sqrt{3}}\ \end{align*} Hence, $CD=\boxed{\textbf{E}}$.

Solution 2 (Cyclic Quadrilaterals, Right Triangles, LoC)

Since $\angle{A}+\angle{C} = 180^{\circ}, ABCD$ is cyclic. Using Ptolemy's Theorem gets $\text{(1): } 4AD + 3CD = \sqrt{37}BD.$

Right triangles $\Delta ABD$ and $\Delta CBD$ obtain $\text{(2): } 9+AD^2=BD^2$ and $\text{(3):} 16+CD^2=BD^2,$ respectively.

Seeing squares in $\text{(2)}$ and $\text{(3)}$, we square $\text{(1)}$ and get $\text{(4): } 16AD^2+9CD^2 + 24AD\cdot CD = 37BD^2.$

We don't like that $AD\cdot CD$ term, but fortunately LoC exists: $37 = AD^2 + CD^2 - 2*AD*CD*\cos(60^{\circ})$. Solving for $AD\cdot CD$ and plugging it into $\text{(4)}$, and using $AD^2 = 7 + CD^2$ and $BD^2 = 16 + CD^2$ from the first two equations, gets $40(CD^2+7)+33CD^2 - 24\cdot 37 = 37(16+CD^2).$

Solve for $CD = \boxed{\textbf{E}}$.

~PureSwag

See also

1992 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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