What is the value of $a + b$ if $asqrt{b} = BC$ in right triangle $Delta ABC$?
$begingroup$
In $Delta ABC$, $angle ABC = 90^circ$ , $D$ is the midpoint of line
$BC$. Point $P$ is on $AD$ line. $PM$ & $PN$ are respectively
perpendicular on $AB$ & $AC$. $PM$ = $2PN$, $AB = 5$, $BC$ = $asqrt{b}$,
where $a, b$ are positive integers. $a+b$ = ?
Source: Bangladesh Math Olympiad 2018 junior category.
I could not find any way to relate $PM$ = $2PN$ to this math.
geometry contest-math triangles
edited Jan 17 at 15:06
the_fox
2,90231538
asked Jan 17 at 13:34
ShromiShromi
3289
$endgroup$
add a comment |
$begingroup$
In $Delta ABC$, $angle ABC = 90^circ$ , $D$ is the midpoint of line
$BC$. Point $P$ is on $AD$ line. $PM$ & $PN$ are respectively
perpendicular on $AB$ & $AC$. $PM$ = $2PN$, $AB = 5$, $BC$ = $asqrt{b}$,
where $a, b$ are positive integers. $a+b$ = ?
Source: Bangladesh Math Olympiad 2018 junior category.
I could not find any way to relate $PM$ = $2PN$ to this math.
geometry contest-math triangles
edited Jan 17 at 15:06
the_fox
2,90231538
asked Jan 17 at 13:34
ShromiShromi
3289
$endgroup$
$begingroup$
This may not be unique. What if $|BC|^2$ were $75$, do we take $a=1$ or $a=5$? Or is the correct value amenable to $a=1$ only?
$endgroup$
– Oscar Lanzi
Jan 17 at 14:18
add a comment |
$begingroup$
In $Delta ABC$, $angle ABC = 90^circ$ , $D$ is the midpoint of line
$BC$. Point $P$ is on $AD$ line. $PM$ & $PN$ are respectively
perpendicular on $AB$ & $AC$. $PM$ = $2PN$, $AB = 5$, $BC$ = $asqrt{b}$,
where $a, b$ are positive integers. $a+b$ = ?
Source: Bangladesh Math Olympiad 2018 junior category.
I could not find any way to relate $PM$ = $2PN$ to this math.
geometry contest-math triangles
edited Jan 17 at 15:06
the_fox
2,90231538
asked Jan 17 at 13:34
ShromiShromi
3289
$endgroup$
In $Delta ABC$, $angle ABC = 90^circ$ , $D$ is the midpoint of line
$BC$. Point $P$ is on $AD$ line. $PM$ & $PN$ are respectively
perpendicular on $AB$ & $AC$. $PM$ = $2PN$, $AB = 5$, $BC$ = $asqrt{b}$,
where $a, b$ are positive integers. $a+b$ = ?
Source: Bangladesh Math Olympiad 2018 junior category.
I could not find any way to relate $PM$ = $2PN$ to this math.
geometry contest-math triangles
geometry contest-math triangles
edited Jan 17 at 15:06
the_fox
2,90231538
asked Jan 17 at 13:34
ShromiShromi
3289
edited Jan 17 at 15:06
the_fox
2,90231538
asked Jan 17 at 13:34
ShromiShromi
3289
edited Jan 17 at 15:06
the_fox
2,90231538
edited Jan 17 at 15:06
the_fox
2,90231538
edited Jan 17 at 15:06
the_fox
2,90231538
2,90231538
asked Jan 17 at 13:34
ShromiShromi
3289
asked Jan 17 at 13:34
ShromiShromi
3289
asked Jan 17 at 13:34
ShromiShromi
3289
3289
$begingroup$
This may not be unique. What if $|BC|^2$ were $75$, do we take $a=1$ or $a=5$? Or is the correct value amenable to $a=1$ only?
$endgroup$
– Oscar Lanzi
Jan 17 at 14:18
add a comment |
$begingroup$
This may not be unique. What if $|BC|^2$ were $75$, do we take $a=1$ or $a=5$? Or is the correct value amenable to $a=1$ only?
$endgroup$
– Oscar Lanzi
Jan 17 at 14:18
$begingroup$
This may not be unique. What if $|BC|^2$ were $75$, do we take $a=1$ or $a=5$? Or is the correct value amenable to $a=1$ only?
$endgroup$
– Oscar Lanzi
Jan 17 at 14:18
$begingroup$
This may not be unique. What if $|BC|^2$ were $75$, do we take $a=1$ or $a=5$? Or is the correct value amenable to $a=1$ only?
$endgroup$
– Oscar Lanzi
Jan 17 at 14:18
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$triangle ABC sim triangle AMM' sim triangle PNM'$
$frac{BC}{AB}=frac{NM'}{PN}$
$PM'=PM=2PN$
$NM'=sqrt{(2PN)^2-PN^2}=PNsqrt3$
$frac{NM'}{PN}=sqrt3$
$frac{BC}{AB}=sqrt3$
$BC=ABsqrt3=5sqrt3$
$a=5,b=3,color{blue}{a+b=8}$
answered Jan 17 at 14:51
Daniel MathiasDaniel Mathias
1,40518
$endgroup$
$begingroup$
That's strange, I thought the sum would be 76.
$endgroup$
– Oscar Lanzi
Jan 17 at 15:02
2
$begingroup$
@OscarLanzi As you pointed out, $5sqrt3=1sqrt{75}$, so $(a,b)=(1,75)$ technically qualifies.
$endgroup$
– Daniel Mathias
Jan 17 at 15:06
add a comment |
$begingroup$
Let $$angle{BAC}=alpha$$ then
$$tan(alpha)=frac{PM}{5-PN}=-frac{PN}{frac{a}{2}-2PN}$$ so we get $$frac{a}{2}+5=3PN$$ (1) and $$AD^2=5^2+frac{a^2}{4}$$ so
$$left(frac{2PN}{sin(alpha)}+frac{PN}{cos(alpha)}right)^2=5^2+frac{a^2}{4}$$ and for $PN$ we have $$PN=frac{1}{3}left(frac{a}{2}+5right)$$ and $$cos^2(alpha)=frac{1}{tan^2(alpha)+1)}$$ and $$sin^2(alpha)=frac{tan^2(alpha)}{1+tan^2(alpha)}$$
Can you finish?
answered Jan 17 at 14:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
add a comment |
$begingroup$
We can think that $B=(0,0)$. Then $A=(0,5)$ and $C=(asqrt{b},0)$.
Then we have that $D=(frac{asqrt{b}}{2},0)$.
Since $P$ is in $AD$, $P=(0,5)+lambda_1 (asqrt{b},-10)$.
Now, as we can see in the picture, we do not know where $N$ and $M$ are. But by what you said, we have two options for each one:
$N_1= (0,5)+lambda_2(asqrt{b},-5)$
$N_2= (0,0)+lambda_3(asqrt{b},0) = (lambda_3 cdot asqrt{b},0)$
$M_1= (0,0)+lambda_4(0,5)=(0,lambda_4 cdot 5)$
$M_2= (0,5)+lambda_5(asqrt{b},-5)$
Now, we have to play computating $PM$ and $PN$ and using your relation.
answered Jan 17 at 13:51
idriskameniidriskameni
749321
$endgroup$
$begingroup$
I will edit it now. Thanks!
$endgroup$
– idriskameni
Jan 17 at 13:55
add a comment |
$begingroup$
Draw the triangle as instructed. $M$ and $N$ are taken to be the feet if the respective perpendiculars to the triangle's sides, $M$ on $AB$ and $N$ on $AC$.
Draw $DN'$ parallel to $PN$ with $N'$ also on $AC$. Then $triangle AMP ~ triangle ABD$ and $triangle APB ~ triangle ADN'$, forcing $|PM|/|PN|=|BD|/|DN'|=|CD|/|DN'|=2$. Then right $triangle DN'C$ has a hypoteneyse/leg ratio of $2:1$ hence $angle C$ measures $30°$.
It follows that right $triangle ABC$ with a $30°$ angle at $C$ also has a $2:1$ ratio of the hypotenuse to the shorter leg, therefore its hypotenuse is $10$ and $|BC|=5sqrt{3}=1sqrt{75}$. And the solution is nonunique! C'mon man!
answered Jan 17 at 14:52
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$triangle ABC sim triangle AMM' sim triangle PNM'$
$frac{BC}{AB}=frac{NM'}{PN}$
$PM'=PM=2PN$
$NM'=sqrt{(2PN)^2-PN^2}=PNsqrt3$
$frac{NM'}{PN}=sqrt3$
$frac{BC}{AB}=sqrt3$
$BC=ABsqrt3=5sqrt3$
$a=5,b=3,color{blue}{a+b=8}$
answered Jan 17 at 14:51
Daniel MathiasDaniel Mathias
1,40518
$endgroup$
$begingroup$
That's strange, I thought the sum would be 76.
$endgroup$
– Oscar Lanzi
Jan 17 at 15:02
2
$begingroup$
@OscarLanzi As you pointed out, $5sqrt3=1sqrt{75}$, so $(a,b)=(1,75)$ technically qualifies.
$endgroup$
– Daniel Mathias
Jan 17 at 15:06
add a comment |
$begingroup$
$triangle ABC sim triangle AMM' sim triangle PNM'$
$frac{BC}{AB}=frac{NM'}{PN}$
$PM'=PM=2PN$
$NM'=sqrt{(2PN)^2-PN^2}=PNsqrt3$
$frac{NM'}{PN}=sqrt3$
$frac{BC}{AB}=sqrt3$
$BC=ABsqrt3=5sqrt3$
$a=5,b=3,color{blue}{a+b=8}$
answered Jan 17 at 14:51
Daniel MathiasDaniel Mathias
1,40518
$endgroup$
$begingroup$
That's strange, I thought the sum would be 76.
$endgroup$
– Oscar Lanzi
Jan 17 at 15:02
2
$begingroup$
@OscarLanzi As you pointed out, $5sqrt3=1sqrt{75}$, so $(a,b)=(1,75)$ technically qualifies.
$endgroup$
– Daniel Mathias
Jan 17 at 15:06
add a comment |
$begingroup$
$triangle ABC sim triangle AMM' sim triangle PNM'$
$frac{BC}{AB}=frac{NM'}{PN}$
$PM'=PM=2PN$
$NM'=sqrt{(2PN)^2-PN^2}=PNsqrt3$
$frac{NM'}{PN}=sqrt3$
$frac{BC}{AB}=sqrt3$
$BC=ABsqrt3=5sqrt3$
$a=5,b=3,color{blue}{a+b=8}$
answered Jan 17 at 14:51
Daniel MathiasDaniel Mathias
1,40518
$endgroup$
$triangle ABC sim triangle AMM' sim triangle PNM'$
$frac{BC}{AB}=frac{NM'}{PN}$
$PM'=PM=2PN$
$NM'=sqrt{(2PN)^2-PN^2}=PNsqrt3$
$frac{NM'}{PN}=sqrt3$
$frac{BC}{AB}=sqrt3$
$BC=ABsqrt3=5sqrt3$
$a=5,b=3,color{blue}{a+b=8}$
answered Jan 17 at 14:51
Daniel MathiasDaniel Mathias
1,40518
answered Jan 17 at 14:51
Daniel MathiasDaniel Mathias
1,40518
answered Jan 17 at 14:51
Daniel MathiasDaniel Mathias
1,40518
answered Jan 17 at 14:51
Daniel MathiasDaniel Mathias
1,40518
1,40518
$begingroup$
That's strange, I thought the sum would be 76.
$endgroup$
– Oscar Lanzi
Jan 17 at 15:02
2
$begingroup$
@OscarLanzi As you pointed out, $5sqrt3=1sqrt{75}$, so $(a,b)=(1,75)$ technically qualifies.
$endgroup$
– Daniel Mathias
Jan 17 at 15:06
add a comment |
$begingroup$
That's strange, I thought the sum would be 76.
$endgroup$
– Oscar Lanzi
Jan 17 at 15:02
2
$begingroup$
@OscarLanzi As you pointed out, $5sqrt3=1sqrt{75}$, so $(a,b)=(1,75)$ technically qualifies.
$endgroup$
– Daniel Mathias
Jan 17 at 15:06
$begingroup$
That's strange, I thought the sum would be 76.
$endgroup$
– Oscar Lanzi
Jan 17 at 15:02
$begingroup$
That's strange, I thought the sum would be 76.
$endgroup$
– Oscar Lanzi
Jan 17 at 15:02
2
2
$begingroup$
@OscarLanzi As you pointed out, $5sqrt3=1sqrt{75}$, so $(a,b)=(1,75)$ technically qualifies.
$endgroup$
– Daniel Mathias
Jan 17 at 15:06
$begingroup$
@OscarLanzi As you pointed out, $5sqrt3=1sqrt{75}$, so $(a,b)=(1,75)$ technically qualifies.
$endgroup$
– Daniel Mathias
Jan 17 at 15:06
add a comment |
$begingroup$
Let $$angle{BAC}=alpha$$ then
$$tan(alpha)=frac{PM}{5-PN}=-frac{PN}{frac{a}{2}-2PN}$$ so we get $$frac{a}{2}+5=3PN$$ (1) and $$AD^2=5^2+frac{a^2}{4}$$ so
$$left(frac{2PN}{sin(alpha)}+frac{PN}{cos(alpha)}right)^2=5^2+frac{a^2}{4}$$ and for $PN$ we have $$PN=frac{1}{3}left(frac{a}{2}+5right)$$ and $$cos^2(alpha)=frac{1}{tan^2(alpha)+1)}$$ and $$sin^2(alpha)=frac{tan^2(alpha)}{1+tan^2(alpha)}$$
Can you finish?
answered Jan 17 at 14:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
add a comment |
$begingroup$
Let $$angle{BAC}=alpha$$ then
$$tan(alpha)=frac{PM}{5-PN}=-frac{PN}{frac{a}{2}-2PN}$$ so we get $$frac{a}{2}+5=3PN$$ (1) and $$AD^2=5^2+frac{a^2}{4}$$ so
$$left(frac{2PN}{sin(alpha)}+frac{PN}{cos(alpha)}right)^2=5^2+frac{a^2}{4}$$ and for $PN$ we have $$PN=frac{1}{3}left(frac{a}{2}+5right)$$ and $$cos^2(alpha)=frac{1}{tan^2(alpha)+1)}$$ and $$sin^2(alpha)=frac{tan^2(alpha)}{1+tan^2(alpha)}$$
Can you finish?
answered Jan 17 at 14:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
add a comment |
$begingroup$
Let $$angle{BAC}=alpha$$ then
$$tan(alpha)=frac{PM}{5-PN}=-frac{PN}{frac{a}{2}-2PN}$$ so we get $$frac{a}{2}+5=3PN$$ (1) and $$AD^2=5^2+frac{a^2}{4}$$ so
$$left(frac{2PN}{sin(alpha)}+frac{PN}{cos(alpha)}right)^2=5^2+frac{a^2}{4}$$ and for $PN$ we have $$PN=frac{1}{3}left(frac{a}{2}+5right)$$ and $$cos^2(alpha)=frac{1}{tan^2(alpha)+1)}$$ and $$sin^2(alpha)=frac{tan^2(alpha)}{1+tan^2(alpha)}$$
Can you finish?
answered Jan 17 at 14:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
Let $$angle{BAC}=alpha$$ then
$$tan(alpha)=frac{PM}{5-PN}=-frac{PN}{frac{a}{2}-2PN}$$ so we get $$frac{a}{2}+5=3PN$$ (1) and $$AD^2=5^2+frac{a^2}{4}$$ so
$$left(frac{2PN}{sin(alpha)}+frac{PN}{cos(alpha)}right)^2=5^2+frac{a^2}{4}$$ and for $PN$ we have $$PN=frac{1}{3}left(frac{a}{2}+5right)$$ and $$cos^2(alpha)=frac{1}{tan^2(alpha)+1)}$$ and $$sin^2(alpha)=frac{tan^2(alpha)}{1+tan^2(alpha)}$$
Can you finish?
answered Jan 17 at 14:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
answered Jan 17 at 14:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
answered Jan 17 at 14:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
answered Jan 17 at 14:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
79k42867
add a comment |
add a comment |
$begingroup$
We can think that $B=(0,0)$. Then $A=(0,5)$ and $C=(asqrt{b},0)$.
Then we have that $D=(frac{asqrt{b}}{2},0)$.
Since $P$ is in $AD$, $P=(0,5)+lambda_1 (asqrt{b},-10)$.
Now, as we can see in the picture, we do not know where $N$ and $M$ are. But by what you said, we have two options for each one:
$N_1= (0,5)+lambda_2(asqrt{b},-5)$
$N_2= (0,0)+lambda_3(asqrt{b},0) = (lambda_3 cdot asqrt{b},0)$
$M_1= (0,0)+lambda_4(0,5)=(0,lambda_4 cdot 5)$
$M_2= (0,5)+lambda_5(asqrt{b},-5)$
Now, we have to play computating $PM$ and $PN$ and using your relation.
answered Jan 17 at 13:51
idriskameniidriskameni
749321
$endgroup$
$begingroup$
I will edit it now. Thanks!
$endgroup$
– idriskameni
Jan 17 at 13:55
add a comment |
$begingroup$
We can think that $B=(0,0)$. Then $A=(0,5)$ and $C=(asqrt{b},0)$.
Then we have that $D=(frac{asqrt{b}}{2},0)$.
Since $P$ is in $AD$, $P=(0,5)+lambda_1 (asqrt{b},-10)$.
Now, as we can see in the picture, we do not know where $N$ and $M$ are. But by what you said, we have two options for each one:
$N_1= (0,5)+lambda_2(asqrt{b},-5)$
$N_2= (0,0)+lambda_3(asqrt{b},0) = (lambda_3 cdot asqrt{b},0)$
$M_1= (0,0)+lambda_4(0,5)=(0,lambda_4 cdot 5)$
$M_2= (0,5)+lambda_5(asqrt{b},-5)$
Now, we have to play computating $PM$ and $PN$ and using your relation.
answered Jan 17 at 13:51
idriskameniidriskameni
749321
$endgroup$
$begingroup$
I will edit it now. Thanks!
$endgroup$
– idriskameni
Jan 17 at 13:55
add a comment |
$begingroup$
We can think that $B=(0,0)$. Then $A=(0,5)$ and $C=(asqrt{b},0)$.
Then we have that $D=(frac{asqrt{b}}{2},0)$.
Since $P$ is in $AD$, $P=(0,5)+lambda_1 (asqrt{b},-10)$.
Now, as we can see in the picture, we do not know where $N$ and $M$ are. But by what you said, we have two options for each one:
$N_1= (0,5)+lambda_2(asqrt{b},-5)$
$N_2= (0,0)+lambda_3(asqrt{b},0) = (lambda_3 cdot asqrt{b},0)$
$M_1= (0,0)+lambda_4(0,5)=(0,lambda_4 cdot 5)$
$M_2= (0,5)+lambda_5(asqrt{b},-5)$
Now, we have to play computating $PM$ and $PN$ and using your relation.
answered Jan 17 at 13:51
idriskameniidriskameni
749321
$endgroup$
We can think that $B=(0,0)$. Then $A=(0,5)$ and $C=(asqrt{b},0)$.
Then we have that $D=(frac{asqrt{b}}{2},0)$.
Since $P$ is in $AD$, $P=(0,5)+lambda_1 (asqrt{b},-10)$.
Now, as we can see in the picture, we do not know where $N$ and $M$ are. But by what you said, we have two options for each one:
$N_1= (0,5)+lambda_2(asqrt{b},-5)$
$N_2= (0,0)+lambda_3(asqrt{b},0) = (lambda_3 cdot asqrt{b},0)$
$M_1= (0,0)+lambda_4(0,5)=(0,lambda_4 cdot 5)$
$M_2= (0,5)+lambda_5(asqrt{b},-5)$
Now, we have to play computating $PM$ and $PN$ and using your relation.
answered Jan 17 at 13:51
idriskameniidriskameni
749321
edited Jan 17 at 14:10
answered Jan 17 at 13:51
idriskameniidriskameni
749321
answered Jan 17 at 13:51
idriskameniidriskameni
749321
answered Jan 17 at 13:51
idriskameniidriskameni
749321
749321
$begingroup$
I will edit it now. Thanks!
$endgroup$
– idriskameni
Jan 17 at 13:55
add a comment |
$begingroup$
I will edit it now. Thanks!
$endgroup$
– idriskameni
Jan 17 at 13:55
$begingroup$
I will edit it now. Thanks!
$endgroup$
– idriskameni
Jan 17 at 13:55
$begingroup$
I will edit it now. Thanks!
$endgroup$
– idriskameni
Jan 17 at 13:55
add a comment |
$begingroup$
Draw the triangle as instructed. $M$ and $N$ are taken to be the feet if the respective perpendiculars to the triangle's sides, $M$ on $AB$ and $N$ on $AC$.
Draw $DN'$ parallel to $PN$ with $N'$ also on $AC$. Then $triangle AMP ~ triangle ABD$ and $triangle APB ~ triangle ADN'$, forcing $|PM|/|PN|=|BD|/|DN'|=|CD|/|DN'|=2$. Then right $triangle DN'C$ has a hypoteneyse/leg ratio of $2:1$ hence $angle C$ measures $30°$.
It follows that right $triangle ABC$ with a $30°$ angle at $C$ also has a $2:1$ ratio of the hypotenuse to the shorter leg, therefore its hypotenuse is $10$ and $|BC|=5sqrt{3}=1sqrt{75}$. And the solution is nonunique! C'mon man!
answered Jan 17 at 14:52
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
add a comment |
$begingroup$
Draw the triangle as instructed. $M$ and $N$ are taken to be the feet if the respective perpendiculars to the triangle's sides, $M$ on $AB$ and $N$ on $AC$.
Draw $DN'$ parallel to $PN$ with $N'$ also on $AC$. Then $triangle AMP ~ triangle ABD$ and $triangle APB ~ triangle ADN'$, forcing $|PM|/|PN|=|BD|/|DN'|=|CD|/|DN'|=2$. Then right $triangle DN'C$ has a hypoteneyse/leg ratio of $2:1$ hence $angle C$ measures $30°$.
It follows that right $triangle ABC$ with a $30°$ angle at $C$ also has a $2:1$ ratio of the hypotenuse to the shorter leg, therefore its hypotenuse is $10$ and $|BC|=5sqrt{3}=1sqrt{75}$. And the solution is nonunique! C'mon man!
answered Jan 17 at 14:52
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
add a comment |
$begingroup$
Draw the triangle as instructed. $M$ and $N$ are taken to be the feet if the respective perpendiculars to the triangle's sides, $M$ on $AB$ and $N$ on $AC$.
Draw $DN'$ parallel to $PN$ with $N'$ also on $AC$. Then $triangle AMP ~ triangle ABD$ and $triangle APB ~ triangle ADN'$, forcing $|PM|/|PN|=|BD|/|DN'|=|CD|/|DN'|=2$. Then right $triangle DN'C$ has a hypoteneyse/leg ratio of $2:1$ hence $angle C$ measures $30°$.
It follows that right $triangle ABC$ with a $30°$ angle at $C$ also has a $2:1$ ratio of the hypotenuse to the shorter leg, therefore its hypotenuse is $10$ and $|BC|=5sqrt{3}=1sqrt{75}$. And the solution is nonunique! C'mon man!
answered Jan 17 at 14:52
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
Draw the triangle as instructed. $M$ and $N$ are taken to be the feet if the respective perpendiculars to the triangle's sides, $M$ on $AB$ and $N$ on $AC$.
Draw $DN'$ parallel to $PN$ with $N'$ also on $AC$. Then $triangle AMP ~ triangle ABD$ and $triangle APB ~ triangle ADN'$, forcing $|PM|/|PN|=|BD|/|DN'|=|CD|/|DN'|=2$. Then right $triangle DN'C$ has a hypoteneyse/leg ratio of $2:1$ hence $angle C$ measures $30°$.
It follows that right $triangle ABC$ with a $30°$ angle at $C$ also has a $2:1$ ratio of the hypotenuse to the shorter leg, therefore its hypotenuse is $10$ and $|BC|=5sqrt{3}=1sqrt{75}$. And the solution is nonunique! C'mon man!
answered Jan 17 at 14:52
Oscar LanziOscar Lanzi
13.6k12136
edited Jan 17 at 14:59
answered Jan 17 at 14:52
Oscar LanziOscar Lanzi
13.6k12136
answered Jan 17 at 14:52
Oscar LanziOscar Lanzi
13.6k12136
answered Jan 17 at 14:52
Oscar LanziOscar Lanzi
13.6k12136
13.6k12136
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$begingroup$
This may not be unique. What if $|BC|^2$ were $75$, do we take $a=1$ or $a=5$? Or is the correct value amenable to $a=1$ only?
$endgroup$
– Oscar Lanzi
Jan 17 at 14:18