AIME 2022 II · 第 12 题

Solution 1 (Sines and Cosines)

Since we are asked to find $\tan \theta$, we can find $\sin \theta$ and $\cos \theta$ separately and use their values to get $\tan \theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$.

We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\sin \theta$ and $\cos \theta$, we can represent these areas using $\sin \theta$ as follows:

$$\begin{aligned} 30 &=[ABCD] \\ &=[AXB] + [BXC] + [CXD] + [DXA] \\ &=\frac{1}{2} ab \sin (\angle AXB) + \frac{1}{2} bc \sin (\angle BXC) + \frac{1}{2} cd \sin (\angle CXD) + \frac{1}{2} da \sin (\angle AXD) \\ &=\frac{1}{2} ab \sin (180^\circ - \theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (180^\circ - \theta) + \frac{1}{2} da \sin (\theta). \end{aligned}$$ We know that $\sin (180^\circ - \theta) = \sin \theta$. Therefore it follows that:

$$\begin{aligned} 30 &=\frac{1}{2} ab \sin (180^\circ - \theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (180^\circ - \theta) + \frac{1}{2} da \sin (\theta) \\ &=\frac{1}{2} ab \sin (\theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (\theta) + \frac{1}{2} da \sin (\theta) \\ &=\frac{1}{2}\sin\theta (ab + bc + cd + da). \end{aligned}$$ From here we see that $\sin \theta = \frac{60}{ab + bc + cd + da}$. Now we need to find $\cos \theta$. Using the Law of Cosines on each of the four smaller triangles, we get following equations:

$$\begin{aligned} 5^2 &= a^2 + b^2 - 2ab\cos(180^\circ-\theta), \\ 6^2 &= b^2 + c^2 - 2bc\cos \theta, \\ 9^2 &= c^2 + d^2 - 2cd\cos(180^\circ-\theta), \\ 7^2 &= d^2 + a^2 - 2da\cos \theta. \end{aligned}$$ We know that $\cos (180^\circ - \theta) = -\cos \theta$ for all $\theta$. We can substitute this value into our equations to get:

$$\begin{aligned} 5^2 &= a^2 + b^2 + 2ab\cos \theta, &&(1) \\ 6^2 &= b^2 + c^2 - 2bc\cos \theta, &&(2) \\ 9^2 &= c^2 + d^2 + 2cd\cos \theta, &&(3) \\ 7^2 &= d^2 + a^2 - 2da\cos \theta. &&(4) \end{aligned}$$ If we subtract $(2)+(4)$ from $(1)+(3)$, the squared terms cancel, leaving us with:

$$\begin{aligned} 5^2 + 9^2 - 6^2 - 7^2 &= 2ab \cos \theta + 2bc \cos \theta + 2cd \cos \theta + 2da \cos \theta \\ 21 &= 2\cos \theta (ab + bc + cd + da). \end{aligned}$$ From here we see that $\cos \theta = \frac{21/2}{ab + bc + cd + da}$.

Since we have figured out $\sin \theta$ and $\cos \theta$, we can calculate $\tan \theta$:

$$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{60}{ab + bc + cd + da}}{\frac{21/2}{ab + bc + cd + da}} = \frac{60}{21/2} = \frac{120}{21} = \frac{40}{7}.$$ Therefore our answer is $40 + 7 = \boxed{047}$.

~ Steven Chen (www.professorchenedu.com)

~ my_aops_lessons

Solution 2 (Right Triangles)

In convex quadrilateral $ABCD,$ let $AB=5,BC=6,CD=9,$ and $DA=7.$ Let $A'$ and $C'$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to $\overline{BD}.$ We obtain the following diagram:

Let $BC'=p,C'E=q,EA'=r,A'D=s,AA'=h_1,$ and $CC'=h_2.$ We apply the Pythagorean Theorem to right triangles $\triangle ABA',\triangle BCC',\triangle CDC',$ and $\triangle DAA',$ respectively:

$$\begin{array}{ccccccccccccccccc} (p+q+r)^2&+&h_1^2&=&5^2, &&&&&&&&&&&&\hspace{36mm}(1) \\ [1ex] p^2&+&h_2^2&=&6^2, &&&&&&&&&&&&\hspace{36mm}(2) \\ [1ex] (q+r+s)^2&+&h_2^2&=&9^2, &&&&&&&&&&&&\hspace{36mm}(3) \\ [1ex] s^2&+&h_1^2&=&7^2. &&&&&&&&&&&&\hspace{36mm}(4) \end{array}$$ Let the brackets denote areas. We get

$$\begin{aligned} [ABD]+[CBD]&=[ABCD] \\ \frac12(p+q+r+s)h_1+\frac12(p+q+r+s)h_2&=30 \\ \frac12(p+q+r+s)(h_1+h_2)&=30 \\ (p+q+r+s)(h_1+h_2)&=60. \hspace{49.25mm}(5) \end{aligned}$$ We subtract $(2)+(4)$ from $(1)+(3):$

$$\begin{aligned} (p+q+r)^2+(q+r+s)^2-p^2-s^2&=21 \\ \left[(p+q+r)^2-s^2\right]+\left[(q+r+s)^2-p^2\right]&=21 \\ (p+q+r+s)(p+q+r-s)+(p+q+r+s)(-p+q+r+s)&=21 \\ (p+q+r+s)(2q+2r)&=21 \\ 2(p+q+r+s)(q+r)&=21 \\ (p+q+r+s)(q+r)&=\frac{21}{2}. \hspace{9.5mm}(6) \end{aligned}$$ From right triangles $\triangle AEA'$ and $\triangle CEC',$ we have $\tan\theta=\frac{h_1}{r}=\frac{h_2}{q}.$ It follows that

$$\begin{alignat*}{8} \tan\theta&=\frac{h_1}{r}\qquad&\implies\qquad h_1&=r\tan\theta, \hspace{64mm}&(1\star)\\ \tan\theta&=\frac{h_2}{q}\qquad&\implies\qquad h_2&=q\tan\theta. &(2\star) \end{alignat*}$$ Finally, we divide $(5)$ by $(6):$

$$\begin{aligned} \frac{h_1+h_2}{q+r}&=\frac{40}{7} \\ \frac{r\tan\theta+q\tan\theta}{q+r}&=\frac{40}{7} \hspace{15mm} &&\text{by }(1\star)\text{ and }(2\star)\\ \frac{(r+q)\tan\theta}{q+r}&=\frac{40}{7} \\ \tan\theta&=\frac{40}{7}, \end{aligned}$$ from which the answer is $40+7=\boxed{047}.$

~MRENTHUSIASM

Solution 3 (Bretschneider's Formula)

Bretschneider's Formula

Given quadrilateral $ABCD$, let, $a, b, c, d$, be the sides, $s$ the semiperimeter, and $u, v$, the diagonals. Then the area, $K$, is given by

$$K = \tfrac 14 \sqrt {4u^2v^2-(b^2+d^2-a^2-c^2)^2}$$

Solution

By Bretschneider's Formula,

$$30=\tfrac{1}{4}\sqrt{4u^2v^2-(b^2+d^2-a^2-c^2)^2}=\tfrac{1}{4}\sqrt{4u^2v^2-441}.$$ Thus, $uv=\tfrac{3\sqrt{1649}}{2}$. Also,

$$[ABCD]=\tfrac 12 \cdot uv\sin{\theta};$$ solving for $\sin{\theta}$ yields $\sin{\theta}=\tfrac{40}{\sqrt{1649}}$. Since $\theta$ is acute, $\cos{\theta}$ is positive, from which $\cos{\theta}=\tfrac{7}{\sqrt{1649}}$. Solving for $\tan{\theta}$ yields

$$\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}=\frac{40}{7},$$ for a final answer of $\boxed{047}$.

~ Leo.Euler

Solution 4 (Symmetry)

Claim

Given an inscribed quadrilateral $ABCD$ with sides $AB = a, BC = b, CD = c,$ and $DA = d.$ Prove that the $\angle \theta < 90^\circ$ between the diagonals is given by

$$\begin{aligned}2(ac + bd) \cos \theta = {|d^2 – c^2 + b^2 – a^2|}.\end{aligned}$$ Proof

Let the point $B'$ be symmetric to $B$ with respect to the perpendicular bisector $AC.$ Then the quadrilateral $AB'CD$ is an inscribed one, $AB' = b, B'C = a.$

$$2 \angle AEB = \overset{\Large\frown} {AB} + \overset{\Large\frown} {CD}.$$ $$\begin{aligned} 2\angle B'AD = \overset{\Large\frown} {B'C} + \overset{\Large\frown} {CD} = \overset{\Large\frown} {AB} + \overset{\Large\frown} {CD} \implies \angle AEB = \angle B'AD.\end{aligned}$$ We apply the Law of Cosines to $\triangle AB'D$ and $\triangle CB'D$:

$$\begin{aligned} B'D^2 = AD^2 + AB'^2 – 2 AD \cdot AB' \cos \theta, \end{aligned}$$ $$\begin{aligned} B'D^2 = CD^2 + CB'^2 + 2 CD \cdot CB' \cos \theta,\end{aligned}$$ $$\begin{aligned} d^2 + b^2 – 2 bd \cos \theta = c^2 + a^2 + 2ac \cos \theta,\end{aligned}$$ $$\begin{aligned} 2(ac + bd) \cos \theta = |d^2 – c^2 + b^2 – a^2|.\end{aligned}$$ vladimir.shelomovskii@gmail.com, vvsss

Note 1

By generalization, the tangent of the acute angle formed by the diagonals is

$$\left|\frac{4A}{a^2-b^2+c^2-d^2}\right|.$$

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