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Difference between revisions of "2021 AIME II Problems/Problem 12"

(→‎Solution 1)
(→‎Solution 1)
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We index this equation as Eq (2).
 
We index this equation as Eq (2).
  
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Analogously, we can establish the following equation for <math>\triangle BEC</math> that <math>b^2 + c^2 - 2 b c \cos \theta = 6^2</math> (indexed as Eq (3)), the following equation for <math>\triangle CED</math> that <math>c^2 + d^2 + 2 c d \cos \theta = 9^2</math> (indexed as Eq (4)) and the following equation for <math>\triangle DEA</math> that <math>d^2 + a^2 - 2 d a \cos \theta = 7^2</math>.
+
Analogously, we can establish the following equation for <math>\triangle BEC</math> that <math>b^2 + c^2 - 2 b c \cos \theta = 6^2</math> (indexed as Eq (3)), the following equation for <math>\triangle CED</math> that <math>c^2 + d^2 + 2 c d \cos \theta = 9^2</math> (indexed as Eq (4)) and the following equation for <math>\triangle DEA</math> that <math>d^2 + a^2 - 2 d a \cos \theta = 7^2</math> (indexed as Eq (5)).
  
−
~ Steven Chen (from Professor Chen Education Palace)
+
By taking Eq (2) - Eq (3) + Eq (4) - Eq (5) and dividing both sides of the equation by 2, we get <math>\left( ab + bc + cd + da \right) \cos \theta = \frac{21}{2}</math>. We index this equation as Eq (6).
 +
 
 +
By taking <math>\frac{Eq (1)}{Eq (6)}</math>, we get <math>\tan \theta = \frac{60}{21/2} = \frac{40}{7}</math>.
 +
 
 +
Therefore, by writing this answer in the form of <math>\frac{m}{n}</math>, we have <math>m = 40</math> and <math>n = 7</math>.
 +
Therefore, the answer to this question is <math>m + n = 40 + 7 = 47</math>.
 +
 
 +
~ Steven Chen (www.professorchenedu.com)
  
 
==See also==
 
==See also==
 
{{AIME box|year=2021|n=II|num-b=11|num-a=13}}
 
{{AIME box|year=2021|n=II|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:50, 22 March 2021

Problem

A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution 1

We denote by $A$, $B$, $C$ and $D$ four vertices of this quadrilateral, such that $AB = 5$, $BC = 6$, $CD = 9$, $DA = 7$. We denote by $E$ the point that two diagonals $AC$ and $BD$ meet at. To simplify the notation, we denote $a = AE$, $b = BE$, $c = CE$, $d = DE$. We denote $\theta = \angle AED$.

First, we use the triangle area formula with sines to write down an equation of the area of the quadrilateral $ABCD$. We have ${\rm Area} \ ABCD = {\rm Area} \ \triangle ABE + {\rm Area} \ \triangle BCE + {\rm Area} \ \triangle CDE + {\rm Area} \ \triangle DAE = \frac{1}{2} ab \sin \angle AEB + \frac{1}{2} bc \sin \angle BEC + \frac{1}{2} cd \sin \angle CED + \frac{1}{2} da \sin \angle DEA = \frac{1}{2} \left( ab + bc + cd + da \right) \sin \theta$.

Because ${\rm Area} \ ABCD = 30$, we have $\left( ab + bc + cd + da \right) \sin \theta = 60$. We index this equation as Eq (1).

Second, we use the law of cosines to establish four equations for four sides of the quadrilateral $ABCD$.

By applying the law of cosines to $\triangle AEB$, we have $a^2 + b^2 - 2 a b \cos \angle AEB = AB^2 = 5^2$. Note that $\cos \angle AEB = \cos \left( 180^\circ - \theta \right) = \cos \theta$. Hence, $a^2 + b^2 + 2 a b \cos \theta = 5^2$. We index this equation as Eq (2).

Analogously, we can establish the following equation for $\triangle BEC$ that $b^2 + c^2 - 2 b c \cos \theta = 6^2$ (indexed as Eq (3)), the following equation for $\triangle CED$ that $c^2 + d^2 + 2 c d \cos \theta = 9^2$ (indexed as Eq (4)) and the following equation for $\triangle DEA$ that $d^2 + a^2 - 2 d a \cos \theta = 7^2$ (indexed as Eq (5)).

By taking Eq (2) - Eq (3) + Eq (4) - Eq (5) and dividing both sides of the equation by 2, we get $\left( ab + bc + cd + da \right) \cos \theta = \frac{21}{2}$. We index this equation as Eq (6).

By taking $\frac{Eq (1)}{Eq (6)}$, we get $\tan \theta = \frac{60}{21/2} = \frac{40}{7}$.

Therefore, by writing this answer in the form of $\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = 47$.

~ Steven Chen (www.professorchenedu.com)

See also

2021 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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