AIME 2019 I · 第 6 题

Solution 1 (Trig)

Let $\angle MKN=\alpha$ and $\angle LNK=\beta$. Let $P$ be the project of $L$ onto line $NK$. Note $\angle KLP=\beta$.

Then, $KP=28\sin\beta=8\cos\alpha$. Furthermore, $KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha$.

Dividing the equations gives

$$\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}$$ Thus, $MK=\frac{MN}{\tan\alpha}=98$, so $MO=MK-KO=\boxed{090}$.

Solution 2 (WRONG SOLUTION Cyclic Quads, PoP)

NOTE: this solution is wrong. The equation is correct due to similar triangles as described in solution 8, not PoP.

Because $\angle KLN = \angle KMN = 90^{\circ}$, $KLMN$ is a cyclic quadrilateral. (THE FOLLOWING SENTENCE IS WRONG) Hence, by Power of Point,

$$KO\cdot KM = KL^2 \implies KM=\dfrac{28^2}{8}=98 \implies MO=98-8=\boxed{090}$$ as desired.

~Mathkiddie

Solution 3 (Similar triangles)

First, let $P$ be the intersection of $LO$ and $KN$ as shown above. Note that $m\angle KPL = 90^{\circ}$ as given in the problem. Since $\angle KPL \cong \angle KLN$ and $\angle PKL \cong \angle LKN$, $\triangle PKL \sim \triangle LKN$ by AA similarity. Similarly, $\triangle KMN \sim \triangle KPO$. Using these similarities we see that

$$\frac{KP}{KL} = \frac{KL}{KN}$$ $$KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}$$ and

$$\frac{KP}{KO} = \frac{KM}{KN}$$ $$KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}$$ Combining the two equations, we get

$$\frac{8\cdot KM}{KN} = \frac{784}{KN}$$ $$8 \cdot KM = 28^2$$ $$KM = 98$$ Since $KM = KO + MO$, we get $MO = 98 -8 = \boxed{090}$.

Solution by vedadehhc

Solution 4 (Similar triangles, orthocenters)

Extend $KL$ and $NM$ past $L$ and $M$ respectively to meet at $P$. Let $H$ be the intersection of diagonals $KM$ and $LN$ (this is the orthocenter of $\triangle KNP$).

As $\triangle KOL \sim \triangle KHP$ (as $LO \parallel PH$, using the fact that $H$ is the orthocenter), we may let $OH = 8k$ and $LP = 28k$.

Then using similarity with triangles $\triangle KLH$ and $\triangle KMP$ we have

$$\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}$$ Cross-multiplying and dividing by $4+4k$ gives $2(8+8k+HM) = 28 \cdot 7 = 196$ so $MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}$. (Solution by scrabbler94)

Solution 5 (Algebraic Bashing)

First, let $P$ be the intersection of $LO$ and $KN$. We can use the right triangles in the problem to create equations. Let $a=NP, b=PK, c=NO, d=OM, e=OP, f=PC,$ and $g=NC.$ We are trying to find $d.$ We can find $7$ equations. They are

$$4225+d^2=c^2,$$ $$4225+d^2+16d+64=a^2+2ab+b^2,$$ $$a^2+e^2=c^2,$$ $$b^2+e^2=64,$$ $$b^2+e^2+2ef+f^2=784,$$ $$a^2+e^2+2ef+f^2=g^2,$$ and

$$g^2+784=a^2+2ab+b^2.$$ We can subtract the fifth equation from the sixth equation to get $a^2-b^2=g^2-784.$ We can subtract the fourth equation from the third equation to get $a^2-b^2=c^2-64.$ Combining these equations gives $c^2-64=g^2-784$ so $g^2=c^2+720.$ Substituting this into the seventh equation gives $c^2+1504=a^2+2ab+b^2.$ Substituting this into the second equation gives $4225+d^2+16d+64=c^2+1504$. Subtracting the first equation from this gives $16d+64=1504.$ Solving this equation, we find that $d=\boxed{090}.$ (Solution by DottedCaculator)

Solution 6 (5-second PoP)

Notice that $KLMN$ is inscribed in the circle with diameter $\overline{KN}$ and $XOMN$ is inscribed in the circle with diameter $\overline{ON}$. Furthermore, $(XLN)$ is tangent to $\overline{KL}$. Then,

$$KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,$$ and $MO=KM-KO=\boxed{090}$.

(Solution by TheUltimate123)

If you're wondering why $KX \cdot KN=KL^2,$ it's because PoP on $(XLN)$ or by $KX \cdot KN=KX \cdot (KX+XN)=KX^2+KX \cdot XN=KX^2+LX^2=KL^2$ (last part by geometric mean theorem / similarity).

Note: the "semicircle" circumscribing points XOMN is not a semicircle. That is just there to tell you that X, O, M, N are indeed concyclic, so ignore the subtlety of the diagram that makes O seems slightly off the marks than it should be.

Solution 7 (Alternative PoP)

(Diagram by vedadehhc)

Call the base of the altitude from $L$ to $NK$ point $P$. Let $PO=x$. Now, we have that $KP=\sqrt{64-x^2}$ by the Pythagorean Theorem. Once again by Pythagorean, $LO=\sqrt{720+x^2}-x$. Using Power of a Point, we have

$$(KO)(OM)=(LO)(OQ)$$ ($Q$ is the intersection of $OL$ with the circle $\neq L$)

$$8(MO)=(\sqrt{720+x^2}-x)(\sqrt{720+x^2}+x)$$ $$8(MO)=720$$ $$MO=\boxed{090}$$ .

(Solution by RootThreeOverTwo)

Remark: Length of OQ

Since $P$ is on the circle’s diameter, $QP = LP = \sqrt{720+x^2}$. So, $OQ = PQ + PO = x + \sqrt{720+x^2}$. ~diyarv

Solution8 (just one pair of similar triangles)

Note that since $\angle KLN = \angle KMN$, quadrilateral $KLMN$ is cyclic. Therefore, we have

$$\angle LMK = \angle LNK = 90^{\circ} - \angle LKN = \angle KLP,$$ so $\triangle KLO \sim \triangle KML$, giving

$$\frac{KM}{28} = \frac{28}{8} \implies KM = 98.$$ Therefore, $OM = 98-8 = \boxed{90}$.

Solution 9 (Pythagoras Bash)

By Pythagorean Theorem, $KM^2+65^2 = KN^2 = 28^2 + LN^2$. Thus, $LN^2 = KM^2 + 65^2 - 28^2$.

By Pythagorean Theorem, $KP^2 + LP^2 = 28^2$, and $PN^2 + LP^2 = LN^2$.

$$PN^2 = (KN - KP)^2 = (\sqrt{KM^2 + 65^2} - KP)^2$$ It follows that

$$(\sqrt{KM^2 + 65^2} - KP)^2 + LP^2 = KM^2 + 65^2 - 28^2$$ $$KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + KP^2 + LP^2 = KM^2 + 65^2 - 28^2$$ Since $KP^2 + LP^2 = 28^2$,

$$KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + 28^2 = KM^2 + 65^2 - 28^2$$ $$-2\sqrt{KM^2 + 65^2}(KP) = -2 \times 28^2$$ $$KP = \frac{28^2}{\sqrt{KM^2 + 65^2}}$$ $\angle OKP = \angle NKM$ (it's the same angle) and $\angle OPK = \angle KMN = 90^{\circ}$. Thus, $\triangle KOP \sim \triangle KNM$.

Thus,

$$\frac{KO}{KN} = \frac{KP}{KM}$$ $$\frac{8}{\sqrt{KM^2 + 65^2}} = \frac{\frac{28^2}{\sqrt{KM^2 + 65^2}}}{KM}$$ Multiplying both sides by $\sqrt{KM^2 + 65^2}$:

$$8 = \frac{28^2}{KM}$$ $$KM = 98$$ Therefore, $OM = 98-8 = \boxed{90}$

~ Solution by adam_zheng

Video Solution

Video Solution: https://www.youtube.com/watch?v=0AXF-5SsLc8

Video Solution 2

https://youtu.be/I-8xZGhoDUY

~Shreyas S

Video Solution 3

https://www.youtube.com/watch?v=pP3cih_8bg4