Wrong Answers Only Pt.2

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MathRook7817

677 posts

MathRook7817

#1 h Apr 9, 2025, 5:49 PM • 2 Y

Y by Exponent11, jkim0656

Problem: What is the area of a triangle with side lengths 13,14, and 15?
WRONG ANSWERS ONLY!

other one got locked for some reason

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Soupboy0

376 posts

Soupboy0

#3 h Apr 9, 2025, 5:54 PM

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$\sqrt{13^2+14^2-15^2} = 80$

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unicornlover9763

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unicornlover9763

#4 h Apr 9, 2025, 5:59 PM

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(13+14)/15=1.8

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Mathdreams

1472 posts

Mathdreams

#5 h Apr 9, 2025, 6:01 PM

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$\sqrt(21 * 8 * 7 * 6) = 1$

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BossLu99

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BossLu99

#7 h Apr 9, 2025, 6:10 PM

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like 6-7

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Soupboy0

376 posts

Soupboy0

#8 h Apr 9, 2025, 6:12 PM

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Using the 1434 Geo Lemma, we find the area of the triangle to be $\frac{[\binom{13}{2}-\binom{13}{4}+\binom{14}{2}-\binom{14}{4}+\binom{15}{2}-\binom{15}{4}][14-15+13]}{13^2+15^2-\frac{14}{2}}(-1)^{\frac{1}{2}(13+14+15)}$

edit: this totally isn't the correct answer

This post has been edited 1 time. Last edited by Soupboy0, Apr 9, 2025, 8:49 PM

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Lankou

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Lankou

#9 h Apr 9, 2025, 6:16 PM

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$13^{14^{15}}\approx 10^{10^{17.24}}$ $

This post has been edited 1 time. Last edited by Lankou, Apr 9, 2025, 6:20 PM

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komodraqon

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komodraqon

#10 h Apr 9, 2025, 6:17 PM

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13+14+15 = 131415

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maxamc

568 posts

maxamc

#11 h Apr 9, 2025, 6:49 PM

Y by

the inradius is $\frac{13 \cdot 14 \cdot 15}{4 Area}$ . it is also $\frac{Area}{\frac{13+14+15}{2}},$ solving gives $Area=\frac{21\sqrt{130}}{2}$ .

This post has been edited 3 times. Last edited by maxamc, Apr 9, 2025, 6:50 PM

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iwastedmyusername

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iwastedmyusername

#12 h Apr 9, 2025, 7:01 PM • 9 Y

Y by komodraqon, donut_bear, Exponent11, MathRook7817, huajun78, aidan0626, legospartan, Rice_Farmer, EthanNg6

We want to find the area of a triangle with side lengths 13, 14, and 15. Now this might seem quite difficult, but there's a neat trick to find the area of this specific triangle. First, split the side with length 14 into two sides of lengths 5 and 9. Now, we want to connect the point connecting the sides with lengths 13 and 15 to the center of the splitting (idk). Hmmm, its not looking promising. Then I realized. What rhymes with five? Hive. Beehive. A beehive is shaped like a bunch of hexagons. Hexagons have 6 sides. So I found that the number 6 has something to do with this. Because its a triangle and triangles have 3 sides, and multiplication is the third operation most students learn in elementary school, I know it has to be 6 times something. What could 6 possibly be multiplied by? Then I realized that I haven't used the 9 yet. So I searched up on google what happened in the year 9. According to Gemini AI which is a very credible source, there was a devastating defeat for the Roman army led by Publiius Quinticlius Varus. Wait. P, Q, V. How many letters are in between Q and V? Let's see, Q, R, S, T, U. There are 4 letters. Because it's a triangle and it has 1 less side than a square, we want to find the square root. So 2. 2 * 6 = 12. Wait a minute. 5-12-13 and 9-12-15 are both right triangles. So I just need to find the area of two right triangles. This is fairly simple. The area of a right triangle is the product of its two legs, divided by 2. So the area of 30+54=84 right? No actually. You see, when you consider the transformations of a function, f(x+b) is actually a shift b to the LEFT of f(x). f(x-b) is a shift to the RIGHT of f(x). Wait a minute. There are two right triangles. And shifting to the right is adding NEGATIVE b. So we have to multiply 84 by -1. So the answer is -84. Q.E.D.

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ChickensEatGrass

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ChickensEatGrass

#14 h Apr 9, 2025, 7:21 PM

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$13^2+14^2=15^2$ using the Pythagorean $- 140$ Theorem, so $13 \cdot 7=91$ .
But, we mustn’t forget to add the $140$ back to make up for lost aura—which I lose every day sob sob sob. Therefore, our final answer is $231$ .

This post has been edited 3 times. Last edited by ChickensEatGrass, Apr 9, 2025, 7:23 PM

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ilikemath247365

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ilikemath247365

#15 h Apr 9, 2025, 7:26 PM

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Remember that $13 + 14 + 15 = 42$ . Now, the area is 4 times the perimeter so $4 * 42 = 168$ .

This post has been edited 1 time. Last edited by ilikemath247365, Apr 9, 2025, 7:26 PM

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Runner1600

12 posts

Runner1600

#16 h Apr 9, 2025, 7:43 PM

Y by

We know that $13=1+1+1+1+1+1+1+1+1+1+1+1+1$ , right and $14=1+1+1+1+1+1+1+1+1+1+1+1+1+1$ and $15=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1$ . So according to the 5th Symphony by Beethoven, you have to plug in any form of adding 1s into the quadratic formula. So $a=1+1+1+1+1+1+1+1+1+1+1+1+1, b=1+1+1+1+1+1+1+1+1+1+1+1+1+1 c=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1$ . Plugging in, we get

$\frac{-(1+1+1+1+1+1+1+1+1+1+1+1+1+1) \pm \sqrt (1+1+1+1+1+1+1+1+1+1+1+1+1+1)^{2}-4(1+1+1+1+1+1+1+1+1+1+1+1+1)(1+1+1+1+1+1+1+1+1+1+1+1+1+1+1)} {2(1+1+1+1+1+1+1+1+1+1+1+1+1)}$

= $\frac{-14 \pm \sqrt (196-780)} {26}$
= $\frac{-7 \pm \sqrt -274} {13}$
So our two solutions are $\frac{-7-\sqrt-274} {13}$ and $\frac{-7+\sqrt-274} {13}$

This post has been edited 2 times. Last edited by Runner1600, Apr 9, 2025, 7:45 PM
Reason: Sillied, forgot about i

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Runner1600

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Runner1600

#17 h Apr 9, 2025, 7:44 PM

Y by

These are the two different areas

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Existing_Human1

211 posts

Existing_Human1

#18 h Apr 9, 2025, 7:46 PM • 1 Y

Y by EthanNg6

Runner1600 wrote:

We know that $13=1+1+1+1+1+1+1+1+1+1+1+1+1$ , right and $14=1+1+1+1+1+1+1+1+1+1+1+1+1+1$ and $15=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1$ . So according to the 5th Symphony by Beethoven, you have to plug in any form of adding 1s into the quadratic formula. So $a=1+1+1+1+1+1+1+1+1+1+1+1+1, b=1+1+1+1+1+1+1+1+1+1+1+1+1+1 c=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1$ . Plugging in, we get

$\frac{-(1+1+1+1+1+1+1+1+1+1+1+1+1+1) \pm \sqrt (1+1+1+1+1+1+1+1+1+1+1+1+1+1)^{2}-4(1+1+1+1+1+1+1+1+1+1+1+1+1)(1+1+1+1+1+1+1+1+1+1+1+1+1+1+1)} {2(1+1+1+1+1+1+1+1+1+1+1+1+1)}$

= $\frac{-14 \pm \sqrt (196-780)} {26}$ = $\frac{-7 \pm -\sqrt 274} {13}$
So our two solutions are $\frac{-7-\sqrt274} {13}$ and $\frac{-7+\sqrt274} {13}$

Wait, I just realized re-listening to Beethoven's 5th Symphony:

Lyrics:
Da da da da
duhn duhn duhn duhn
My name is Beethoven
And what you should do
Is, whenever you have to find the area of a 13-14-15 triangle, plug the numbers into the quadratic formula
Da da da da

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Soupboy0

376 posts

Soupboy0

#62 h Apr 10, 2025, 2:55 AM • 1 Y

Y by legospartan

WARNING: WHAT IM ABOUT TO GO THROUGH INVOLVES A LENGTHY COORDBASH. IF ONE IS SENSITIVE TO SUCH CONTENT, I WOULD STRONGLY RECOMMEND TO USE THE 1434 GEO LEMMA INSTEAD. ONE NOT FAMILLIAR WITH THE 1434 GEO LEMMA CAN READ MY EARLIER SOLUTION. PLEASE DO NOT THAT THE 2 FORMULAS USED IN THIS TEXT ARE USEFUL FOR COMPETITIONS LIKE THE AMC14 AND THE AMY.THE AREA MUST FIRST BE FOUND IN TERMS OF THE ARBITRARY VARIABLE $T$ . NOTE THAT $T^4 \equiv 1\pmod 4$ , AND THE AREA OF THE TRIANGLE CAN BE FOUND USING 1434 GEO LEMMA OR THE METHOD IM ABOUT TO DESCRIBE. FIRST, DEFINE THE SKIBIDI POINT $S(P) = (x, y)$ OF A TRIANGLE WITH A POINT $P$ AT CARTESIAN COORDINATES $(x, y)$ TRANSFORMED TO $(x-y+5, x+y)$ , AND THEN THE DISTANCE FROM THE ORIGIN IS FOUND. NOW USE THE chicken jockey theorem TO FIND THAT THE AREA OF THE TRIANGLE IN TERMS OF THE ARBITRARY VALUE $T$ IS $\frac{1}{3}(36-T^3)(T^3+1)$ . NOW, WE USE THE SKIBIDI POINT OF OUR TRIANGLE AND LET THE VERTEX OPPOSITE THE SIDE OF LENGTH $15$ REST AT $A=(0, 0)$ SO THAT THE VERTEX OPPOSITE THE SIDE OF LENGTH $14$ IS LOCATED AT $B=(5, 12)$ , AND THE LAST VERTEX IS AT $C=(0, 14)$ . NOW, WE USE THE chicken jockey lemma WHICH STATES THAT $\text{JACK BLACK'S AGE} + [S(C) \cdot S(B)]^2 - ([S(B)]^2-1])([S(C)]^2-1) - 5S(A) = T$ . THIS ULTIMATELY EVALUATES TO $T=3$ , SO PLUGGING $T=3$ INTO OUR EARILER FORMULA YEALDS $\frac{1}{3}(9)(28)=\fbox{84}$ . HOWEVER, THE ULTIMATE FAULT OF THE chicken jockey lemma LIES IN THE FACT THAT IF $S(P) = 5$ IN ANY TRIANGLE, THEN WE MUST MULTIPLY BY $\frac{T^5-T-1}{12}$ BECAUSE OF LOST AREA, AND IN OUR TRIANGLE, $A=(0, 0)$ AND $S(A) = (0-0+5, 0+0) = (5, 0) = 5$ . THEREFORE, THE ANSWER IS $84 \cdot \frac{243-3-1}{14} = \frac{84 \cdot 239}{14} = 6 \cdot 239 = 1434$ . THEREFORE, THIS PROVES THAT JACK BLACK IS STEVE.

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Yiyj1

1266 posts

Yiyj1

#63 h Apr 10, 2025, 3:09 AM

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What is 1434*9+1434+9

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Yrock

1289 posts

Yrock

#64 h Apr 10, 2025, 3:29 AM

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Using Heron's formula:
$s=\frac{13+14+15}{2}=21$
$A=\sqrt{21\cdot(21-13)\cdot(21-14)\cdot(21-15)}=\sqrt{21\cdot8\cdot7\cdot6}=\sqrt{7\cdot7\cdot3\cdot3\cdot8\cdot2}=\sqrt{(7\cdot3\cdot4)^2}=7\cdot3\cdot4 = \boxed{1434}$ .

hehehe

This post has been edited 1 time. Last edited by Yrock, Apr 10, 2025, 3:29 AM

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Soupboy0

376 posts

Soupboy0

#65 h Apr 10, 2025, 3:38 AM

Y by

Yrock wrote:

Using Heron's formula:
$s=\frac{13+14+15}{2}=21$
$A=\sqrt{21\cdot(21-13)\cdot(21-14)\cdot(21-15)}=\sqrt{21\cdot8\cdot7\cdot6}=\sqrt{7\cdot7\cdot3\cdot3\cdot8\cdot2}=\sqrt{(7\cdot3\cdot4)^2}=7\cdot3\cdot4 = \boxed{1434}$ .

hehehe

makes sense although you forgot to use the 1434 geo lemma

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Yrock

1289 posts

Yrock

#66 h Apr 10, 2025, 3:41 AM

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Soupboy0 wrote:

Yrock wrote:

Using Heron's formula:
$s=\frac{13+14+15}{2}=21$
$A=\sqrt{21\cdot(21-13)\cdot(21-14)\cdot(21-15)}=\sqrt{21\cdot8\cdot7\cdot6}=\sqrt{7\cdot7\cdot3\cdot3\cdot8\cdot2}=\sqrt{(7\cdot3\cdot4)^2}=7\cdot3\cdot4 = \boxed{1434}$ .

hehehe

makes sense although you forgot to use the 1434 geo lemma

The fact that every number is equal to 1434 is the 1434 trivial lemma.

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maxamc

568 posts

maxamc

#68 h Apr 10, 2025, 3:46 AM

Y by

EaZ_Shadow wrote:

Triangle has base 14, height of 12, so the answer is 84. $\phantom{01}$

EaZ_Shadow is a genius and clearly should win a fields medal for this correct statement.

Attachments:

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HacheB2031

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HacheB2031

#69 h Apr 10, 2025, 4:49 AM

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HacheB2031 wrote:

the area is $2$
proof: look at the diagram
$[asy] unitsize(1cm); pair A = (0,0); pair B = (2,0); pair C = (0,2); draw(A--B--C--cycle); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NW); label("$14$",A--C,W); label("$15$",B--C,NE); label("$13$",A--B,S); label("$2$",(1,1),2SW); [/asy]$
the diagram is true and stuff so yeah the area is $2$ because the diagram said so

erm actually my diagram is better and proves the area is $3$ and stuff by like the induction hypothesis and the really weird lemma and the proof of the theorem and the because i said so strategy and like a really complicated proof and i bolded this i italiciziaeds edsdedsd this ~~i strikethroughed this~~ i underlined this WHY IS THIS CAPITAL AND STUFF ITS IMPORTANT so yeah my argument is true
$[asy] unitsize(1cm); pair A = (0,0); pair B = (2,0); pair C = (0,3); draw(A--B--C--cycle); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NW); label("$14$",A--C,W); label("$15$",B--C,NE); label("$13$",A--B,S); label("$3$",(1,1.5),2SW); [/asy]$

This post has been edited 1 time. Last edited by HacheB2031, Apr 10, 2025, 4:56 AM

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mdk2013

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mdk2013

#70 h Apr 10, 2025, 4:57 AM

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i said this last time and i say it now: if the side lengths are 13, 14, 15 by the pywagorean teorem we get that the mass of the sun is equal to the mass of jupiter * 0.5647

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HacheB2031

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HacheB2031

#71 h Apr 10, 2025, 4:59 AM • 1 Y

Y by legospartan

The 1434 Theorem: Let $f:\mathbb R\to\mathbb R$ be any real valued function. Then $f(x)=1434.$
"Because I Said So" Theorem: Let $1$ and $2$ be real numbers such that $1=1$ and $2=2.$ Then $1=2.$
Proof: If this wasn't true, then we would die. So, it must be true. Q.E.D.

Proof of 1434: By the "Because I Said So" theorem, the theorem is true. Q.E.D.

Corollary: the area of the triangle is $1434$

This post has been edited 3 times. Last edited by HacheB2031, Apr 10, 2025, 5:02 AM

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jkim0656

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jkim0656

#72 h Apr 10, 2025, 5:00 AM

Y by

side len 13, 14, 15?
seems like a job for the jkim formula!
$\dfrac{\dfrac{13}{14}}{15+13+14}$
= $\dfrac{13}{588}$ ~
that is our totally correct answer

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whwlqkd

99 posts

whwlqkd

#73 h Apr 10, 2025, 7:57 AM

Y by

The triangle has length $13cm,14cm,15mm$
So we get $sqrt{14.25\times 0.25\times 1.25 \times 12.75}=\frac{3}{16} \sqrt{1615}$ .

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kjhgyuio

59 posts

kjhgyuio

#74 h Apr 10, 2025, 10:46 AM

Y by

MathRook7817 wrote:

Problem: What is the area of a triangle with side lengths 13,14, and 15?
WRONG ANSWERS ONLY!

other one got locked for some reason

7

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fruitmonster97

2491 posts

fruitmonster97

#75 h Apr 10, 2025, 11:40 AM • 2 Y

Y by legospartan, ChaitraliKA

We use the well known area=2*perimeter theorem to get $2(13+14+15)=84$ - oh darn, you want a wrong answer

We can split it into two right triangles. the area of a right triangle is simply the sum of the largest and smallest sides times two. This yields 2(9+15)+2(5+13)=84 - whoops mb

We use bird formula. Birds are cool. the number 6 is also cool. being the middle child is cool. Thus 6 times the middle side 14 equals 84 - oh wow im bad at this.

xooks xonks area=1434-iltg by 1434 geo lemma. basic computation yields i=2(13+14)/3=18. l=15. t=14-13=1. g is the shortest side made from splitting into two pythag triples, or 5. Our answer is 1434-(15)(18)(1)(5)=84 - oops sry

~~lamelo ball is 6' 7''. he plays like hes short, so this triangle wants to be double 6' 7'', or 2(6)(7)=84~~ - well this is awkward

the height is 12. thus, ans is 15*14-13*12+15+14+13-12=84 - ok i give up

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Aaronjudgeisgoat

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Aaronjudgeisgoat

#76 h Apr 10, 2025, 1:02 PM

Y by

fruitmonster97 wrote:

We use the well known area=2*perimeter theorem to get $2(13+14+15)=84$ - oh darn, you want a wrong answer

We can split it into two right triangles. the area of a right triangle is simply the sum of the largest and smallest sides times two. This yields 2(9+15)+2(5+13)=84 - whoops mb

We use bird formula. Birds are cool. the number 6 is also cool. being the middle child is cool. Thus 6 times the middle side 14 equals 84 - oh wow im bad at this.

xooks xonks area=1434-iltg by 1434 geo lemma. basic computation yields i=2(13+14)/3=18. l=15. t=14-13=1. g is the shortest side made from splitting into two pythag triples, or 5. Our answer is 1434-(15)(18)(1)(5)=84 - oops sry

~~lamelo ball is 6' 7''. he plays like hes short, so this triangle wants to be double 6' 7'', or 2(6)(7)=84~~ - well this is awkward

the height is 12. thus, ans is 15*14-13*12+15+14+13-12=84 - ok i give up

larry bird ahh moment (iykyk)

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MathRook7817

677 posts

MathRook7817

#77 h Apr 10, 2025, 4:13 PM • 1 Y

Y by ZMB038

The (in)correct answer is:

Area = 3/(1/13+1/14+1/15)

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