AIME 2024 I · 第 5 题

Solution 1

We use simple geometry to solve this problem.

We are given that $A$, $D$, $H$, and $G$ are concyclic; call the circle that they all pass through circle $\omega$ with center $O$. We know that, given any chord on a circle, the perpendicular bisector to the chord passes through the center; thus, given two chords, taking the intersection of their perpendicular bisectors gives the center. We therefore consider chords $HG$ and $AD$ and take the midpoints of $HG$ and $AD$ to be $P$ and $Q$, respectively.

We could draw the circumcircle, but actually it does not matter for our solution; all that matters is that $OA=OH=r$, where $r$ is the circumradius.

By the Pythagorean Theorem, $OQ^2+QA^2=OA^2$. Also, $OP^2+PH^2=OH^2$. We know that $OQ=DE+HP$, and $HP=\dfrac{184}2=92$; $QA=\dfrac{16}2=8$; $OP=DQ+HE=8+17=25$; and finally, $PH=92$. Let $DE=x$. We now know that $OA^2=(x+92)^2+8^2$ and $OH^2=25^2+92^2$. Recall that $OA=OH$; thus, $OA^2=OH^2$. We solve for $x$:

\begin{align*} (x+92)^2+8^2&=25^2+92^2 \\ (x+92)^2&=625+(100-8)^2-8^2 \\ &=625+10000-1600+64-64 \\ &=9025 \\ x+92&=95 \\ x&=3. \\ \end{align*}

The question asks for $CE$, which is $CD-x=107-3=\boxed{104}$.

~Technodoggo

Solution 2

Suppose $DE=x$. Extend $AD$ and $GH$ until they meet at $P$. From the Power of a Point Theorem, we have $(PH)(PG)=(PD)(PA)$. Substituting in these values, we get $(x)(x+184)=(17)(33)=561$. We can use guess and check to find that $x=3$, so $EC=\boxed{104}$.

~alexanderruan

~diagram by Technodoggo

rabbit47 - quadratic actually factors as (x+187)(x-3)=0, from which x=3

Solution 3

We find that

$$\angle GAB = 90-\angle DAG = 90 - (180 - \angle GHD) = \angle DHE.$$ Let $x = DE$ and $T = FG \cap AB$. By similar triangles $\triangle DHE \sim \triangle GAT$ we have $\frac{DE}{EH} = \frac{GT}{AT}$. Substituting lengths we have $\frac{x}{17} = \frac{16 + 17}{184 + x}.$ Solving, we find $x = 3$ and thus $CE = 107 - 3 = \boxed{104}.$ ~AtharvNaphade ~coolruler ~eevee9406

Solution 4

One liner: $107-\sqrt{92^2+25^2-8^2}+92=\boxed{104}$

~Bluesoul

Explanation

Let $OP$ intersect $DF$ at $T$ (using the same diagram as Solution 2).

The formula calculates the distance from $O$ to $H$ (or $G$), $\sqrt{92^2+25^2}$, then shifts it to $OD$ and the finds the distance from $O$ to $Q$, $\sqrt{92^2+25^2-8^2}$. $107$ minus that gives $CT$, and when added to $92$, half of $FE=TE$, gives $CT+TE=CE$

Solution 5

Let $\angle{DHE} = \theta.$ This means that $DE = 17\tan{\theta}.$ Since quadrilateral $ADHG$ is cyclic, $\angle{DAG} = 180 - \angle{DHG} = 90 - \theta.$

Let $X = AG \cap DF.$ Then, $\Delta DXA \sim \Delta FXG,$ with side ratio $16:17.$ Also, since $\angle{DAG} = 90 - \theta, \angle{DXA} = \angle{FXG} = \theta.$ Using the similar triangles, we have $\tan{\theta} = \frac{16}{DX} = \frac{17}{FX}$ and $DX + FX = DE + EF = 17\tan{\theta} + 184.$

Since we want $CE = CD - DE = 107 - 17\tan{\theta},$ we only need to solve for $\tan{\theta}$ in this system of equations. Solving yields $\tan{\theta} = \frac{3}{17},$ so $CE = \boxed{104.}$

~PureSwag

Solution 6

Using a ruler (also acting as a straight edge), draw the figure to scale with one unit = 1mm. With a compass, draw circles until you get one such that $A,D,H,G$ are on the edge of the drawn circle. From here, measuring with your ruler should give $CE = \boxed{104.}$

Note: 1 mm is probably the best unit to use here just for convenience (drawing all required parts of the figure fits into a normal-sized scrap paper 8.5 x 11); also all lines can be drawn with a standard 12-inch ruler

~kipper

Solution 7(Ptolemy's Theorem)

Since ADHG is cyclic, AH*DG=AD*HG+DH*AG,let DE be x, we represent the equation in terms of x,and you get $x=3, 107-3= \boxed{104.}$

Ps: extremely time consuming, do not use on the test

Video Solution with Circle Properties

https://youtu.be/1LWwJeFpU9Y ~Veer Mahajan

Video Solution 1 by OmegaLearn.org

https://youtu.be/Ss-u5auH4fE

Video Solution 2

https://youtu.be/R6dkIKuZHsM

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Fast Video Solution by Do Math or Go Home

https://www.youtube.com/watch?v=Hz3PGY_a9Hc