AIME 2014 I · 第 15 题

Solution 1

Since $\angle DBE = 90^\circ$, $DE$ is the diameter of $\omega$. Then $\angle DFE=\angle DGE=90^\circ$. But $DF=FE$, so $\triangle DEF$ is a 45-45-90 triangle. Letting $DG=3x$, we have that $EG=4x$, $DE=5x$, and $DF=EF=\frac{5x}{\sqrt{2}}$.

Note that $\triangle DGE \sim \triangle ABC$ by SAS similarity, so $\angle BAC = \angle GDE$ and $\angle ACB = \angle DEG$. Since $DEFG$ is a cyclic quadrilateral, $\angle BAC = \angle GDE=180^\circ-\angle EFG = \angle AFE$ and $\angle ACB = \angle DEG = \angle GFD$, implying that $\triangle AFE$ and $\triangle CDF$ are isosceles. As a result, $AE=CD=\frac{5x}{\sqrt{2}}$, so $BE=3-\frac{5x}{\sqrt{2}}$ and $BD =4-\frac{5x}{\sqrt{2}}$.

Finally, using the Pythagorean Theorem on $\triangle BDE$,

$$\left(3-\frac{5x}{\sqrt{2}}\right)^2 + \left(4-\frac{5x}{\sqrt{2}}\right)^2 = (5x)^2$$ Solving for $x$, we get that $x=\frac{5\sqrt{2}}{14}$, so $DE= 5x =\frac{25 \sqrt{2}}{14}$. Thus, the answer is $25+2+14=\boxed{041}$.

Solution 2

First we note that $\triangle DEF$ is an isosceles right triangle with hypotenuse $\overline{DE}$ the same as the diameter of $\omega$. We also note that $\triangle DGE \sim \triangle ABC$ since $\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$.

From congruent arc intersections, we know that $\angle GED \cong \angle GBC$, and that from similar triangles $\angle GED$ is also congruent to $\angle GCB$. Thus, $\triangle BGC$ is an isosceles triangle with $BG = GC$, so $G$ is the midpoint of $\overline{AC}$ and $AG = GC = 5/2$. Similarly, we can find from angle chasing that $\angle ABF = \angle EDF = \frac{\pi}4$. Therefore, $\overline{BF}$ is the angle bisector of $\angle B$. From the angle bisector theorem, we have $\frac{AF}{AB} = \frac{CF}{CB}$, so $AF = 15/7$ and $CF = 20/7$.

Lastly, we apply power of a point from points $A$ and $C$ with respect to $\omega$ and have $AE \times AB=AF \times AG$ and $CD \times CB=CG \times CF$, so we can compute that $EB = \frac{17}{14}$ and $DB = \frac{31}{14}$. From the Pythagorean Theorem, we result in $DE = \frac{25 \sqrt{2}}{14}$, so $a+b+c=25+2+14= \boxed{041}$

Also: $FG=\frac{20}{7}-\frac{5}{2}=\frac{5}{2}-\frac{15}{7}=\frac{5}{14}$. We can also use Ptolemy's Theorem on quadrilateral $DEFG$ to figure what $FG$ is in terms of $d$:

$$DE\cdot FG+DG\cdot EF=DF\cdot EG$$ $$d\cdot FG+\frac{3d}{5}\cdot \frac{d}{\sqrt{2}}=\frac{4d}{5}\cdot \frac{d}{\sqrt{2}}$$ $$d\cdot FG+\frac{3d^2}{5\sqrt{2}}=\frac{4d^2}{5\sqrt{2}}\implies FG=\frac{d}{5\sqrt{2}}$$ Thus $\frac{d}{5\sqrt{2}}=\frac{5}{14}\rightarrow d=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}$. $a+b+c=25+2+14= \boxed{041}$

Solution 3

Call $DE=x$ and as a result $DF=EF=\frac{x\sqrt{2}}{2}, EG=\frac{4x}{5}, GD=\frac{3x}{5}$. Since $EFGD$ is cyclic we just need to get $DG$ and using LoS(for more detail see the $2$nd paragraph of Solution $2$) we get $AG=\frac{5}{2}$ and using a similar argument(use LoS again) and subtracting you get $FG=\frac{5}{14}$ so you can use Ptolemy to get $x=\frac{25\sqrt{2}}{14} \implies \boxed{041}$. ~First

Solution 4

See inside the $\triangle DEF$, we can find that $AG>AF$ since if $AG, we can see that Ptolemy Theorem inside cyclic quadrilateral$EFGD$doesn't work. Now let's see when$AG>AF$, since$\frac{DG}{EG} = \frac{3}{4}$, we can assume that$EG=4x;GD=3x;ED=5x$, since we know$EF=FD$so$\triangle EFD$is isosceles right triangle. We can denote$DF=EF=\frac{5x\sqrt{2}}{2}$.Applying Ptolemy Theorem inside the cyclic quadrilateral$EFGD$we can get the length of$FG$can be represented as$\frac{x\sqrt{2}}{2}$. After observing, we can see$\angle AFE=\angle EDG$, whereas$\angle A=\angle EDG$so we can see$\triangle AEF$is isosceles triangle. Since$\triangle ABC$is a$3-4-5$triangle so we can directly know that the length of AF can be written in the form of$3x\sqrt{2}$. Denoting a point$J$on side$AC$with that$DJ$is perpendicular to side$AC$. Now with the same reason, we can see that$\triangle DJG$is a isosceles right triangle, so we can get$GJ=\frac{3x\sqrt{2}}{2}$while the segment$CJ$is$2x\sqrt{2}$since its 3-4-5 again. Now adding all those segments together we can find that$AC=5=7x\sqrt{2}$and$x=\frac{5\sqrt{2}}{14}$and the desired$ED=5x=\frac{25\sqrt{2}}{14}$which our answer is$\boxed{041}$ ~bluesoul

Solution 5

The main element of the solution is the proof that $BF$ is bisector of $\angle B.$

Let $O$ be the midpoint of $DE.$ $\angle EBF = 90^\circ \implies$

$O$ is the center of the circle $BDGFE.$ $\angle EOF = 90^\circ \implies \overset{\Large\frown} {EF} = 90^\circ \implies \angle EBF = 45^\circ \implies$ BF is bisector of $\angle ABC\implies BF = \frac {2AB \cdot BC}{AB+BC} \cos 45^\circ =\frac {12 \cdot \sqrt{2}}{7}.$

$$\angle EGD = 90^\circ, \frac {EG}{GD}=\frac{4}{3} \implies$$ $$\angle GED = \angle GCD =\gamma \implies \overset{\Large\frown} {DG} = 2\gamma.$$ $$2\angle ACB = \overset{\Large\frown} {BEF} - \overset{\Large\frown} {DG} \implies \overset{\Large\frown} {BEF} = 4 \gamma \implies$$ $$\angle BOF = 4 \gamma \implies \angle OBF = \angle OFB = 90^\circ – 2 \gamma.$$ Let $BO = EO = DO = r \implies BF = 2 r \cos(90^\circ – 2\gamma) =$

$$=2 r \sin 2\gamma = 4r \sin \gamma \cdot \cos \gamma = 4 r\cdot \frac {3}{5} \cdot \frac {4}{5} = \frac {48}{25} = \frac {12 \cdot \sqrt{2}}{7}\implies$$ $$r = \frac {25 \cdot \sqrt{2}}{28}\implies ED = 2r = \frac {25 \cdot \sqrt{2}}{14}\implies \boxed{\textbf{041}}.$$ vladimir.shelomovskii@gmail.com, vvsss

Solution 6

The main element of the solution is the proof that $G$ is midpoint of $AC.$

As in Solution 5 we get $\angle GED = \angle DBG =\gamma \implies$

$\triangle BCG$ is isosceles triangle with $BG=CG.$

Similarly $BG = AG \implies AG = CG = BG = \frac {AC}{2} =\frac {5}{2}.$

$$\overset{\Large\frown} {FG} = 90^\circ – \overset{\Large\frown} {GD} = 90^\circ – 2\gamma \implies$$ $$\overset{\Large\frown} {BFG} = 4\gamma + 90^\circ – 2\gamma = 90^\circ + 2\gamma \implies$$ $$\angle BOG = 90^\circ + 2\gamma \implies \angle BGO = \angle GBO = 45^\circ - \gamma.$$ Let $\hspace{10mm} BO = EO = DO = r \implies$

$$BG = 2 r \cos(45^\circ – \gamma) = 2 r (\sin \gamma + \cos \gamma)\frac {\sqrt {2}}{2} =$$ $$r \biggl(\frac {3}{5} + \frac {4}{5}\biggr) \sqrt {2} = r \frac {7 \sqrt{2}}{5} = \frac {5}{2}\implies$$ $$r = \frac {25 \cdot \sqrt{2}}{28}\implies ED = \frac {25 \cdot \sqrt{2}}{14}\implies \boxed{\textbf{041}}.$$ vladimir.shelomovskii@gmail.com, vvsss

Solution 7

Let $(BEFGD) = \omega$. By Incenter-Excenter(Fact $5$), $F$ is the angle bisector of $\angle B$. Then by Ratio Lemma we have

$$\frac{AG}{CG} = \frac{\sin(ABG)}{\sin(CBG)} \cdot \frac{AB}{BC} = \frac{\sin(GDE)}{\sin(DEG)} \cdot \frac{3}{4} = 1$$ Thus, $G$ is the midpoint of $AC$.

We can calculate $AF$ and $CF$ to be $\frac{15}{7}$ and $\frac{20}{7}$ respectively. And then by Power of a Point, we have $\newline$

$$\operatorname{Pow}_{\omega}(A) = AE \cdot AB = AF \cdot AG \implies AE = \frac{25}{14}$$ And then similarly, we have $CD = AE = \frac{25}{14}$. $\newline$

Then $EB = \frac{17}{14}$ and $DB = \frac{31}{14}$ and by Pythagorean we have $DE = \frac{25\sqrt{2}}{14}$, so our answer is $\boxed{\textbf{041}}.$

~dolphinday

Solution 8 (funny angle chase & trig)

Since $\angle EBD$ is right, $DE$ is clearly the diameter. Let $EF=DF=x, ED=x\sqrt{2}, DG=\tfrac{3x\sqrt{2}}{5}, EG=\tfrac{4x\sqrt{2}}{5}$. Then, let $\angle DEG=\alpha$. Therefore, $\angle EFG=\angle EFD+\angle DFG=90^{\circ}+\alpha,$ and $\angle EFA=90^{\circ}-\alpha$. However, $\triangle DGE \sim \triangle ABC$ so $\angle AFE$ also equals $90^{\circ}-\alpha$. Thus, $\triangle AEF$ is isosceles with $\angle FAE \cong \angle EFA \implies AE=EF=x.$

Furthermore, $\angle FEG = \angle FDG = \angle EDG-\angle EDF=90^{\circ}-\alpha-45^{\circ}=45^{\circ}-\alpha$. Also, $\angle AEF=2\alpha$ in triangle $\triangle AEF$, thus $\angle BED=135^{\circ}-2\alpha$ since $\angle AEB=180^{\circ}$. Using $\cos \alpha=\tfrac{4}{5}$, it's relatively easy to derive that $\cos (135^{\circ}-2\alpha)=\tfrac{17\sqrt{2}}{50}$. Since $\cos(135^{\circ}-2\alpha)=\tfrac{BE}{DE}$, we get that $BE=\tfrac{17x}{25}$. Finally, since $AE+BE=x+\tfrac{17x}{25}=3$, we solve for

$$x=\tfrac{25}{14} \implies DE=x\sqrt{2}=\tfrac{25\sqrt{2}}{14},$$ so our desired answer is $\boxed{041}.$

~SirAppel

Solution 9 (Vectors)

$\angle{B} = 90^\circ ==>$ $ED$ is the diameter of the circle $==> \angle{EGD} = 90^\circ$

Because $\frac{GD}{GE} = \frac{AB}{BC} = \frac{3}{4}, \Delta DGE \sim \Delta ABC$

$\angle{DFC} = \angle{DEG} = \angle{C} ==> CD = DF = EF$

We flip the triangle for easier calculation and let $C$ be the origin.

Let $x = CD$

$DF = x\cdot e^{i2\alpha}, EF = x\cdot e^{-(90^\circ-2\alpha)}$

$E$ is on $AB$, which means that $Re(E)$ = 4.

$$Re(x+x\cdot e^{i2\alpha}+x\cdot e^{-(90^\circ-2\alpha)}) = 4$$ $$Re(x+x(\cos{2\alpha}+i\sin{2\alpha})+x(\cos{(2\alpha-90^\circ)}+i\sin{(2\alpha-90^\circ)})) = 4$$ $$x(1+\cos{2\alpha}+\sin{2\alpha})=4$$ $\cos{2\alpha} = \cos^2{\alpha} - \sin^2{\alpha} = \frac{7}{25}$

$\sin{2\alpha} = 2\sin{\alpha}\cos{\alpha} = \frac{24}{25}$

$$x\cdot \frac{56}{25} = 4$$ $$x = \frac{25}{14}$$ Therefore $DE=\sqrt{2}\cdot \frac{25}{14} = \frac{25\sqrt{2}}{14}$

$25+2+14=\boxed{041}$

~cassphe

Solution 10(Very similar to Solution 1 but with an added LoS)

As noted in Solution 1, $DE$ is in fact the diameter of the circle. The proof is trivial. Note that $\triangle ABC$ is a right triangle which means $\angle ABC = \angle EBD = 90^{\circ}$. Therefore $\triangle EBD$ is an inscribed right triangle of the circle which means its arc length is just double the inscribed angle(following the Inscribed Angle Theorem) meaning the circle is cut in half by $DE \implies DE$ is indeed the diameter of the circle.

Similarly, as $DE$ is the diameter, inscribed angles $\angle EFD = \angle EGD = 90^{\circ}$. As $EF = DF$, it follows $\angle FED = \angle FDE = 45^{\circ} \implies \triangle EGD$ is isosceles.

Let $EF = DF = y$:

$ED = y\sqrt{2}$

Now as $\frac{DG}{EG} = \frac{3}{4} \implies DG = 3x, EG = 4x$

Hence $\triangle EGD$ is a right triangle with legs of length $3x, 4x$ and hypotenuse of length $y\sqrt{2}$. Obviously, by $SAS$ similarity, $\triangle EGD \sim \triangle ABC \implies \frac{DG}{ED} = \frac{AB}{AC} \implies \frac{3x}{y\sqrt{2}} = \frac{3}{5} \implies y\sqrt{2} = 5x \implies y = \frac{5x}{\sqrt{2}}$.

Now note that quadrilateral $EFGD$ is cyclic which yields opposite angles are supplementary. Hence

$\angle FED + \angle FGD = 180^{\circ}$ and $\angle FGD + \angle DGC = 180^{\circ}$ following Angle Addition Postulate. Hence by substitution, $\angle FED = \angle DGC = 45^{\circ}$.

Let $\angle ACB = \theta$. Hence $\sin(\theta) = \frac{3}{5}$ following the $3-4-5$ right triangle $ABC$. Now $\angle ACB = \angle GCD = \theta$ so we can apply Law of Sines on $\triangle DGC$.

We already know $\angle DGC = 45^{\circ}$. Also $\angle GCD = \theta$ and $GD = 3x$. Let $DC = a$. Now we apply Law of Sines:

$\frac{DC}{\sin(\angle DGC)} = \frac{GD}{\sin(\angle GCD)} \implies \frac{a}{\sin(45^{\circ})} = \frac{3x}{\sin(\theta)} \implies \frac{a}{\frac{\sqrt{2}}{2}} = \frac{3x}{\frac{3}{5}} \implies a = \frac{5x}{\sqrt{2}}$. This means $DC = \frac{5x}{\sqrt{2}}$.

Now based on how we found $\angle FED = \angle DGC = 45^{\circ}$ using a combination of Supplementary Angle Theorem for cyclic quadrilaterals and Angle Addition Postulate, we can conclude based on a similar logic that $\angle EDG = \angle EFA$. The proof is below:

$\angle EDG + \angle EFG = 180^{\circ}$ and $\angle EFG + \angle EFA = 180^{\circ}$. By substitution, $\angle EDG = \angle EFA$.

Now let $\angle EDG = \alpha$. From the right triangle $\triangle EDG$, $\sin(\angle EDG) = \frac{EG}{ED} = \frac{4x}{5x} = \frac{4}{5}$. So $\sin(\alpha) = \frac{4}{5}$.

Now let $\angle BAC = \angle EAF = \beta$. From the right triangle $\triangle ABC$, $\sin(\angle BAC) = \frac{4}{5} \implies \sin(\angle EAF) = \frac{4}{5} \implies \sin(\beta) = \frac{4}{5}$.

Now we apply Law of Sines on $\triangle EAF$:

$\frac{EF}{\sin(\angle EAF)} = \frac{AE}{\sin(\angle EFA)} \implies \frac{\frac{5x}{\sqrt{2}}}{\sin(\angle \beta)} = \frac{AE}{\sin(\alpha)} \implies \frac{\frac{5x}{\sqrt{2}}}{\frac{4}{5}} = \frac{AE}{\frac{4}{5}} \implies AE = \frac{5x}{\sqrt{2}}$.

Hence $EB = AB - AE = 3 - \frac{5x}{\sqrt{2}}$ and $BD = CB - CD = 4 - \frac{5x}{\sqrt{2}}$.

Now since $\triangle EBD$ is a right triangle with legs of $EB, BD$ and hypotenuse $ED = 5x$ we solve

$(3 - \frac{5x}{\sqrt{2}})^{2} + (4 - \frac{5x}{\sqrt{2}})^{2} = (5x)^{2}$ $9 - 6 \cdot \frac{5x}{\sqrt{2}} + \frac{25x^{2}}{2} + 16 - 8 \cdot \frac{5x}{\sqrt{2}} + \frac{25x^{2}}{2} = 25x^{2}$ $25 - \frac{30x}{\sqrt{2}} - \frac{40x}{\sqrt{2}} = 0$

$25 = \frac{70x}{\sqrt{2}} = 35x\sqrt{2}$.

Now recall we're looking for $ED = 5x$ so we divide both sides by $7\sqrt{2}$

$\frac{25}{7\sqrt{2}} = 5x \implies 5x = \frac{25\sqrt{2}}{14}$.

Thus the answer is $25 + 2 + 14 = \boxed{41}$.

~ilikemath247365

Video Solution by mop 2024

https://youtu.be/GxxZYZrQl2A

~r00tsOfUnity