They mixed up USAJMO and AIME I guess

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Math4Life7

1703 posts

Math4Life7

#1 h Mar 20, 2024, 4:00 AM • 1 Y

Y by Rounak_iitr

Let $ABCD$ be a cyclic quadrilateral with $AB = 7$ and $CD = 8$ . Point $P$ and $Q$ are selected on segment $AB$ such that $AP = BQ = 3$ . Points $R$ and $S$ are selected on segment $CD$ such that $CR = DS = 2$ . Prove that $PQRS$ is a cyclic quadrilateral.

Proposed by Evan O'Dorney

This post has been edited 3 times. Last edited by Math4Life7, Mar 22, 2024, 4:08 PM

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TheHazard

93 posts

TheHazard

#2 h Mar 20, 2024, 4:00 AM • 4 Y

Y by sixoneeight, pikapika007, khina, Spiritpalm

Let $O$ be the center of $(ABCD)$ with radius $R$ . Clearly $OP = OQ$ and $OR = OS$ . I claim that in fact $O$ is the center of $PQRS$ . Let $X$ and $Y$ be the feet from $O$ to $\overline{AB}$ and $\overline{CD}$ . Note that $X$ , $Y$ are the midpoints of $\overline{PQ}$ and $\overline{RS}$ . Now
$\begin{align*} OX^2 &= R^2 - (3.5)^2 \implies OP^2 = XP^2 + OX^2 = (0.5)^2 + R^2 - (3.5)^2 = R^2 - 12 \\ OY^2 &= R^2 - 4^2 \implies OR^2 = YR^2 + OY^2 = 2^2 + R^2 - 4^2 = R^2 - 12 \end{align*}$ so $OP = OR$ . Similarly $OQ = OS$ , done.

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Math4Life7

1703 posts

Math4Life7

#3 h Mar 20, 2024, 4:00 AM • 1 Y

Y by Vladimir_Djurica

Let $O$ be the center of $(ABCD)$ which has radius $R$ . Let $x$ and $y$ be the distance of the perpendicular line from $O$ to $AB$ and $CD$ respectively. We can see that $4^2 + y^2 = R^2 = \left(\frac 72 \right)^2 + x^2 \Rightarrow y^2 + \frac{15}{4} = x^2.$ We can see that $O$ lies on the perpendicular bisector of both $PQ$ and $RS$ . We just need to prove that $OP = OS$ or that $x^2 + \left(\frac 12 \right)^2 = y^2 + 2^2$ which is evident. $\blacksquare$

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Technodoggo

1928 posts

Technodoggo

#4 h Mar 20, 2024, 4:00 AM • 1 Y

Y by Spiritpalm

First, let $E$ and $F$ be the midpoints of $AB$ and $CD$ , respectively. It is clear that $AE=BE=3.5$ , $PE=QE=0.5$ , $DF=CF=4$ , and $SF=RF=2$ . Also, let $O$ be the circumcenter of $ABCD$ .

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; /* image dimensions */ pen wrwrwr = rgb(0.38,0.38,0.38); /* draw figures */ draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr); draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr); draw((7.59,-6.88)--(-0.29,-8.22), linewidth(2) + wrwrwr); draw((3.46,2.59)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((-0.29,-8.22)--(-2.52,-1.01), linewidth(2) + wrwrwr); draw((0.03,0.52)--(1.67,-7.89), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr); /* dots and labels */ dot((2.92,-3.28),dotstyle); label("$O$", (2.43,-3.56), NE * labelscalefactor); dot((-2.52,-1.01),dotstyle); label("$A$", (-2.91,-0.91), NE * labelscalefactor); dot((3.46,2.59),linewidth(4pt) + dotstyle); label("$B$", (3.49,2.78), NE * labelscalefactor); dot((7.59,-6.88),dotstyle); label("$C$", (7.82,-7.24), NE * labelscalefactor); dot((-0.29,-8.22),linewidth(4pt) + dotstyle); label("$D$", (-0.53,-8.62), NE * labelscalefactor); dot((0.03,0.52),linewidth(4pt) + dotstyle); label("$P$", (-0.13,0.67), NE * labelscalefactor); dot((0.89,1.04),linewidth(4pt) + dotstyle); label("$Q$", (0.62,1.16), NE * labelscalefactor); dot((5.61,-7.22),linewidth(4pt) + dotstyle); label("$R$", (5.70,-7.05), NE * labelscalefactor); dot((1.67,-7.89),linewidth(4pt) + dotstyle); label("$S$", (1.75,-7.73), NE * labelscalefactor); dot((0.46,0.78),linewidth(4pt) + dotstyle); label("$E$", (0.26,0.93), NE * labelscalefactor); dot((3.64,-7.55),linewidth(4pt) + dotstyle); label("$F$", (3.73,-7.39), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

By properties of cyclic quadrilaterals, we know that the circumcenter of a cyclic quadrilateral is the intersection of its sides' perpendicular bisectors. This implies that $OE\perp AB$ and $OF\perp CD$ . Since $E$ and $F$ are also bisectors of $PQ$ and $RS$ , respectively, if $PQRS$ is indeed a cyclic quadrilateral, then its circumcenter is also at $O$ . Thus, it suffices to show that $OP=OQ=OR=OS$ .

Notice that $PE=QE$ , $EO=EO$ , and $\angle QEO=\angle PEO=90^\circ$ . By SAS congruency, $\Delta QOE\cong\Delta POE\implies QO=PO$ . Similarly, we find that $\Delta SOF\cong\Delta ROF$ and $OS=OR$ . We now need only to show that these two pairs are equal to each other.

Draw the segments connecting $O$ to $B$ , $Q$ , $C$ , and $R$ .

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; /* image dimensions */ pen wrwrwr = rgb(0.38,0.38,0.38); /* draw figures */ draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr); draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr); draw((7.59,-6.88)--(-0.29,-8.22), linewidth(2) + wrwrwr); draw((3.46,2.59)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((-0.29,-8.22)--(-2.52,-1.01), linewidth(2) + wrwrwr); draw((0.03,0.52)--(1.67,-7.89), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr); draw((0.46,0.78)--(2.92,-3.28), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(3.64,-7.55), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(2.92,-3.28), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(3.46,2.59), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(0.89,1.04), linewidth(2) + wrwrwr); /* dots and labels */ dot((2.92,-3.28),dotstyle); label("$O$", (2.43,-3.56), NE * labelscalefactor); dot((-2.52,-1.01),dotstyle); label("$A$", (-2.91,-0.91), NE * labelscalefactor); dot((3.46,2.59),linewidth(1pt) + dotstyle); label("$B$", (3.49,2.78), NE * labelscalefactor); dot((7.59,-6.88),dotstyle); label("$C$", (7.82,-7.24), NE * labelscalefactor); dot((-0.29,-8.22),linewidth(1pt) + dotstyle); label("$D$", (-0.53,-8.62), NE * labelscalefactor); dot((0.03,0.52),linewidth(1pt) + dotstyle); label("$P$", (-0.13,0.67), NE * labelscalefactor); dot((0.89,1.04),linewidth(1pt) + dotstyle); label("$Q$", (0.62,1.16), NE * labelscalefactor); dot((5.61,-7.22),linewidth(1pt) + dotstyle); label("$R$", (5.70,-7.05), NE * labelscalefactor); dot((1.67,-7.89),linewidth(1pt) + dotstyle); label("$S$", (1.75,-7.73), NE * labelscalefactor); dot((0.46,0.78),linewidth(1pt) + dotstyle); label("$E$", (0.26,0.93), NE * labelscalefactor); dot((3.64,-7.55),linewidth(1pt) + dotstyle); label("$F$", (3.73,-7.39), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

Also, let $r$ be the circumradius of $ABCD$ . This means that $AO=BO=CO=DO=r$ . Recall that $\angle BEO=90^\circ$ and $\angle CFO=90^\circ$ . Notice the several right triangles in our figure.

Let us apply Pythagorean Theorem on $\Delta BEO$ . We can see that $EO^2+EB^2=BO^2\implies EO^2+3.5^2=r^2\implies EO=\sqrt{r^2-12.25}.$

Let us again apply Pythagorean Theorem on $\Delta QEO$ . We can see that $QE^2+EO^2=QO^2\implies0.5^2+r^2-12.25=QO^2\implies QO=\sqrt{r^2-12}.$

Let us apply Pythagorean Theorem on $\Delta CFO$ . We get $CF^2+OF^2=OC^2\implies4^2+OF^2=r^2\implies OF=\sqrt{r^2-16}$ .

We finally apply Pythagorean Theorem on $\Delta RFO$ . This becomes $OF^2+FR^2=OR^2\implies r^2-16+2^2=OR^2\implies OR=\sqrt{r^2-12}$ .

This is the same expression as we got for $QO$ . Thus, $OQ=OR$ , and recalling that $OQ=OP$ and $OR=OS$ , we have shown that $OP=OQ=OR=OS$ . We are done. QED

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rhydon516

531 posts

rhydon516

#5 h Mar 20, 2024, 4:01 AM

Y by

only solve of day 1 oops

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Bluesoul

871 posts

Bluesoul

#6 h Mar 20, 2024, 4:01 AM • 1 Y

Y by JingheZhang

Denote the center of $(ABCD)$ as $O$ , the radius of $(ABCD)$ as $R$ . As the problem states, we have $OP^2=R^2-(\frac{7}{2})^2+(\frac{1}{2})^2=R^2-12=OQ^2; OS^2=R^2-4^2+2^2=R^2-12=OR^2=OP^2=OQ^2$ . Thus, $P,Q,R,S$ are concyclic with center $O$ .

Aime problem 1 more likely

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meth4life2020

28 posts

meth4life2020

#7 h Mar 20, 2024, 4:02 AM

Y by

why maa, why.

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shendrew7

792 posts

shendrew7

#8 h Mar 20, 2024, 4:02 AM • 2 Y

Y by fura3334, a_smart_alecks

Two approaches, which probably constitute upwards of 95% of submitted solutions.

Perpendicular Bisectors: Let $O$ , $R$ denote the center and radius of $(ABCD)$ . Drawing the perpendicular bisectors of $AB$ and $CD$ , we find that $O$ is simply the center of $(PQRS)$ , as
$OP = OQ = R^2 - \left(\frac 72\right)^2 + \left(\frac 12\right)^2 = R^2 - 4^2 + 2^2 = OR = OS. \quad \blacksquare$
Power of a Point: If $AB \parallel CD$ , notice $PQRS$ is an isosceles trapezoid. Otherwise, consider the power of $AB \cap CD$ , and conclude by noting
$x(x+7) = y(y+8) \iff (x+3)(x+4) = (y+2)(y+6). \quad \blacksquare$

This post has been edited 1 time. Last edited by shendrew7, Mar 20, 2024, 4:34 AM

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sixoneeight

1129 posts

sixoneeight

#9 h Mar 20, 2024, 4:02 AM • 3 Y

Y by mathfan2020, fura3334, Jack_w

Headsolved in 1 minute. This must have came from the AMC 8 Shortlist

Solution: Note that the perpendicular bisectors of the chords are the same as the perpendicular bisectors of $PQ$ and $RS$ , so if the claim is true, then the original circle's center must be the circumcenter. We only need to show that $OP = OQ = OR = OS$ , which is equivalent to $(R^2-3.5^2)+0.5^2 = (R^2-4^2)+2^2$ , which is true.

This post has been edited 1 time. Last edited by sixoneeight, Mar 20, 2024, 4:12 AM

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gracemoon124

872 posts

gracemoon124

#10 h Mar 20, 2024, 4:06 AM

Y by

extend + intersect + pop kills

only solve of day 1

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AlexWin0806

50 posts

AlexWin0806

#11 h Mar 20, 2024, 4:31 AM • 2 Y

Y by aidan0626, LLL2019

for PoP solution people:
don't forget AB parallel to CD configuration
there's no intersection when extended

instead it's just isosceles trapezoid ---> cyclic

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OlympusHero

17018 posts

OlympusHero

#12 h Mar 20, 2024, 4:35 AM

Y by

This problem is completely ridiculous.

Drawing the circumcircle of $ABCD$ with center $O$ and the perpendicular bisectors to each side reveals that the perpendicular bisector from $O$ to $AB$ is also the perpendicular bisector from $O$ to $PQ$ , and the perpendicular bisector from $O$ to $BC$ is also the perpendicular bisector from $O$ to $RS$ . It remains to show these two lengths are equal, which can be done with simple length chasing. Thus $PQRS$ is cyclic, done.

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LearnMath_105

130 posts

LearnMath_105

#13 h Mar 20, 2024, 4:36 AM

Y by

AlexWin0806 wrote:

for PoP solution people:
don't forget AB parallel to CD configuration
there's no intersection when extended

instead it's just isosceles trapezoid ---> cyclic

how many points docked if you didnt include it?

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mathboy282

2987 posts

mathboy282

#15 h Mar 20, 2024, 4:46 AM

Y by

AlexWin0806 wrote:

for PoP solution people:
don't forget AB parallel to CD configuration
there's no intersection when extended

instead it's just isosceles trapezoid ---> cyclic

Oh boy.

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the_mathmagician

467 posts

the_mathmagician

#16 h Mar 20, 2024, 5:03 AM

Y by

I literally just started my journey into oly and head-solve-sketch-ed it... I missed JMO slightly but it's good to know I could've kept my "nonzero score" promise pretty easy

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lovematch13

652 posts

lovematch13

#47 h Mar 24, 2024, 7:01 PM

Y by

KevinYang2.71 wrote:

thought process during problem:
1. its not barybashable
2. tried to complex bash. i forgot all the formulas and stuff so i rederived cylic quad condition. ended up with no conjugates stuff. i dont know how to complex bash facepalm
3. found easy pythag sol

relatable

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joshualiu315

2513 posts

joshualiu315

#48 h Mar 24, 2024, 7:17 PM

Y by

Notice that we can use Pythagorean Theorem to show that $OP^2=OQ^2=OR^2=OS^2=r^2-12$ , where $r$ is the circumradius of $(ABCD)$ .

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a_smart_alecks

51 posts

a_smart_alecks

#49 h Mar 24, 2024, 11:56 PM • 1 Y

Y by Jack_w

Glad to see that Evan also found the PoP solution to be more natural than the unintuitive and inferior Pythagorean Theorem solution, which requires testers to observe the difficult equality $4^2-2^2 = 3.5^2-0.5^2$ .

This post has been edited 1 time. Last edited by a_smart_alecks, Mar 24, 2024, 11:56 PM

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oinava

506 posts

oinava

#50 h Mar 26, 2024, 9:32 PM

Y by

Naive power of a point for $P$ and $R$ along diameters works too (to get that the distance to circle and therefore distance to center is the same for both sides' endpoints) without even knowing (essentially, we're proving) the theorem that having a common power implies they form a concentric circle.

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Jishnu4414l

154 posts

Jishnu4414l

#51 h Mar 30, 2024, 2:22 PM

Y by

For God's sake is this AMC 8?
Solution

This post has been edited 1 time. Last edited by Jishnu4414l, Mar 30, 2024, 2:23 PM

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cj13609517288

1867 posts

cj13609517288

#52 h Mar 30, 2024, 5:57 PM • 1 Y

Y by ehuseyinyigit

Let $\omega$ be the circumcircle of quadrilateral $ABCD$ . Then the powers of $P,Q,R,S$ with respect to $\omega$ are all $-12$ , so they are concyclic on a circle concentric with $\omega$ .

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RedFireTruck

4216 posts

RedFireTruck

#53 h Apr 10, 2024, 6:18 PM • 1 Y

Y by ihatemath123

Trivial by PoP. $a(a+7)=b(b+8)$ implies $(a+3)(a+4)=(b+2)(b+6)$ .

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Mathsboy100

135 posts

Mathsboy100

#54 h Oct 1, 2024, 1:42 PM • 1 Y

Y by ihatemath123

So can you guys help me out with the Nigerian math olympiad 2023 it's missing

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ehuseyinyigit

777 posts

ehuseyinyigit

#55 h Oct 19, 2024, 10:03 PM

Y by

USAJMO used to have some more general problems, not computational right?
$-------------------------------------------------------$
Let $AB\cap CD=T$ . Suppose that $TB=a$ and $TC=b$ . Then $a(a+7)=b(b+8)$ which implies $(a+3)(a+4)=(b+2)(b+6)$ . The rest follows easily by Reverse PoP as desired.

This post has been edited 2 times. Last edited by ehuseyinyigit, Oct 19, 2024, 10:04 PM

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blueprimes

303 posts

blueprimes

#56 h Yesterday at 12:23 AM • 4 Y

Y by eg4334, aliz, littlefox_amc, aidan0626

We claim that the power of $P, Q, R, S,$ with respect to $(ABCD)$ are equal, implying they are equidistant from the center of $(ABCD)$ which readily implies $PQRS$ is cyclic. The power of $P$ and $Q$ are $AP \cdot PB = AQ \cdot QB = 4 \times 3 = 3 \times 4$ by the commutative property of multiplication. On the other hand, we have $CR \cdot RD = CS \cdot SD = 2 \times 6 = 6 \times 2$ by the commutative property of multiplication again. It suffices to show the miraculous fact
$3 \times 4 = 6 \times 2.$ Computing this out normally would take an amount of time far too long in the span of 4.5 hours, so we cleverly use the distributive property:
$(1 + 1 + 1)(1 + 1 + 1 + 1) = (1 + 1 + 1 + 1 + 1 + 1)(1 + 1).$ Note that $1 \times 1 = 1$ , so expanding the above we want to show that
$1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1.$ Counting with our fingers, there are an equal number of $1$ on both sides, so we may conclude. $QED$

EDIT: Unfortunately, eg4334 pointed out that this is a fakesolve due to several major assumptions:
- Humans only have 10 fingers
- We are assuming $1 \times 1 = 1$
- Assumed distributive property
I will have to carefully retrace my steps to write a proper solution. I am almost at the "Put all fingers down and count again after you reach 10" chapter in EGMO, so hopefully a complete solution will come out soon.

This post has been edited 2 times. Last edited by blueprimes, Yesterday at 12:28 AM

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littlefox_amc

14 posts

littlefox_amc

#57 h Yesterday at 12:36 AM • 4 Y

Y by megarnie, blueprimes, eg4334, Alex-131

blueprimes wrote:

Unfortunately, eg4334 pointed out that this is a fakesolve due to several major assumptions:
- Humans only have 10 fingers
- We are assuming $1 \times 1 = 1$
- Assumed distributive property
I will have to carefully retrace my steps to write a proper solution. I am almost at the "Put all fingers down and count again after you reach 10" chapter in EGMO, so hopefully a complete solution will come out soon.

dw i will help patch up the fakesolve
1. Use pencils
2. Sorry this is an unsolved problem in mathematics. but you should be able to quote it without proof
3. idk what the distributive property is. You can make a dot diagram and count with pencils though

edit: count again after you reach 10?? without pencils?!!?!

This post has been edited 1 time. Last edited by littlefox_amc, Yesterday at 12:40 AM

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Ilikeminecraft

298 posts

Ilikeminecraft

#58 h Yesterday at 7:28 PM

Y by

note that the power of $P, Q, R, S$ are all 12. since the radii is equal, we can use the distance formula for pop to show that the distance from $O$ to $P,Q,R,S$ are all equal

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maxamc

446 posts

maxamc

#59 h Yesterday at 9:02 PM

Y by

blueprimes wrote:

We claim that the power of $P, Q, R, S,$ with respect to $(ABCD)$ are equal, implying they are equidistant from the center of $(ABCD)$ which readily implies $PQRS$ is cyclic. The power of $P$ and $Q$ are $AP \cdot PB = AQ \cdot QB = 4 \times 3 = 3 \times 4$ by the commutative property of multiplication. On the other hand, we have $CR \cdot RD = CS \cdot SD = 2 \times 6 = 6 \times 2$ by the commutative property of multiplication again. It suffices to show the miraculous fact
$3 \times 4 = 6 \times 2.$ Computing this out normally would take an amount of time far too long in the span of 4.5 hours, so we cleverly use the distributive property:
$(1 + 1 + 1)(1 + 1 + 1 + 1) = (1 + 1 + 1 + 1 + 1 + 1)(1 + 1).$ Note that $1 \times 1 = 1$ , so expanding the above we want to show that
$1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1.$ Counting with our fingers, there are an equal number of $1$ on both sides, so we may conclude. $QED$

EDIT: Unfortunately, eg4334 pointed out that this is a fakesolve due to several major assumptions:
- Humans only have 10 fingers
- We are assuming $1 \times 1 = 1$
- Assumed distributive property
I will have to carefully retrace my steps to write a proper solution. I am almost at the "Put all fingers down and count again after you reach 10" chapter in EGMO, so hopefully a complete solution will come out soon.

To prove the miraculous fact
$3 \times 4 = 6 \times 2,$ We use the Peano axioms and the definition of addition and multiplication in the natural numbers.

$3 \times 4 = 3 \times S(3) = 3 + 3 \times 3 = 3 + 3 \times S(2) = 3 + 3 + 3 \times 2 = 3 + 3 + 3 \times S(1) = 3 + 3 + 3 + 3 \times 1 = 3 + 3 + 3 + 3 \times S(0) = 3 + 3 + 3 + 3 + 3 \times 0 = 3 + 3 + 3 + 3 = 3 + 3 + S(S(S(3))) = 3 + 3 + 6 = 3 + S(S(S(S(S(S(3)))))) = 3 + 9 = S(S(S(S(S(S(S(S(S(3))))))))) = 12.$

$2 \times 6 = 6 \times 2 = 6 \times S(1) = 6 + 6 \times 1 = 6 + 6 \times S(0) = 6 + 6 + 6 \times 0 = 6 + 6 = S(S(S(S(S(S(6)))))) = 12$ .

Since

$3 \times 4 = 12,$
$6 \times 2 = 12,$
$12=12,$
$3 \times 4 = 6 \times 2.$
$\blacksquare$

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